Q 


MECHANICAL  DRAWING. 


PREPARED  FOR  THE  USE  OF  THE  STUDENTS 
OF  THE 

MASSACHUSETTS  INSTITUTE  OF  TECHNOLOGY, 

BOSTON,  MASS. 


LINUS  FAUNCE. 


BOSTON : 
W.  J.  Schofield,  Printer,  105  Summer  Street. 

1887. 


Copyrighted,  1887,  by  Linus  Faunce. 


CONTENTS. 


Page. 
CHAPTER  I. —  Instruments  and  their  Uses 5 

II. —  Geometrical  Problems 13 

III. —  Inking 55 

"        Tinting 61 

IV. —  Projections 65 

"        Notation 68 

"        Projections  of  Straight  Lines 69 

"  "  "    Surfaces 75 

"  "    Solids 80 

V. —  Shadows .'    .     .     .     96 

VI. —  Isometrical  Drawing 112 

"       Oblique  Projections 120 

VII. —  Working  Drawings 121 

VIII.— Examples 127 


2066025 


MECHANICAL  DRAWING. 


CHAPTER  I. 

INSTRUMENTS    AND    THEIR    USES. 

1.  To  do  good  work,  good  instruments  are  essential.  An 
accomplished  draftsman  may  do  fair  work  with  poor  instruments, 
but  the  beginner  will  find  it  sufficiently  hard  to  do  creditable 
work  without  being  handicapped  by  poor  instruments.  It  is 
also  essential  that  the  instruments  should  be  kept  in  good  order. 
They  should  be  handled  carefully,  and  wiped,  before  being  put 
away,  with  wash-leather  or  chamois  skin.  This  is  especially 
needful  if  the  hands  perspire  perceptibly. 

2.  Pencilling.  Drawings  should  always  be  first  made  in 
pencil,  and  inked  afterwards  if  desired.  The  idea  of  pencilling 
is  to  locate  the  lines  exactly,  and  to  make  them  of  the  required 
length.  Accuracy  in  a  drawing  can  only  be  obtained  by  accu- 
racy in  the  pencil  construction.  There  is  a  great  tendency 
among  beginners  to  overlook  this  important  fact,  and  to  become 
careless  in  pencilling,  thinking  they  will  be  able  to  correct  their 
inaccuracies  when  inking.     This  is  a  ereat  mistake,  and  one  to 


6  INSTRUMENTS   AND   THEIR   USES. 

be  especially  avoided.  This  accuracy  can  be  obtained  in  pencil 
only  by  making  very  Jine,  light  lines,  and  to  this  end  hard  pen- 
cils, 6  H,  should  be  used,  and  they  should  be  kept  sharp.  For 
drawing  straight  lines  the  pencil  should  be  sharpened  to  a 
flat,  thin  edge,  like  a  wedge.  The  compass  pencils  should  be 
sharpened  to  a  point.  A  softer  pencil,  4  H,  sharpened  to  a 
point,  should  be  used  in  making  letters,  figures,  etc. 

It  should  be  borne  in  mind  that  a  6  H  pencil  sharpened  to  a 
chisel  point  will  make  a  depression  in  the  paper,  which  can 
never  be  erased,  if  much  pressure  is  put  on  the  pencil ;  hence, 
press  very  lightly  when  using  a  hard  pencil,  so  as  to  avoid  this 
difficulty.  If  a  drawing  is  not  to  be  inked,  but  made  for  rough 
use  in  the  shop,  or  where  accuracy  of  construction  (in  the  draw- 
ing) is  not  essentia],  or  to  be  traced,  a  soft  pencil  would  prefer- 
ably be  used,  the  lines  being  made  somewhat  thicker,  or  heavier. 

3.  Compasses.  In  using  the  compasses  the  lower  part  of 
the  legs  should  be  kept  nearly  vertical,  so  that  the  needle  point 
will  make  only  a  small  hole  in  revolving,  and  both  nibs  of  the 
pen  may  press  equally  on  the  paper.  In  pencilling  it  is  not  so 
essential  that  the  pencil  point  be  kept  vertical,  but  it  is  well  to 
learn  to  use  them  in  one  way,  whether  pencilling  or  inking. 

Hold  the  compasses  loosely  between  the  thumb  and  fore- 
finger only,  and  do  not  press  the  needle  point  into  the  paper. 
If  it  is  sharp,  as  it  should  be,  the  weight  of  the  compass  will  be 
sufficient  to  keep  it  in  place.  While  revolving,  lean  the  com- 
pass very  slightly  in  the  direction  of  revolution,  and  put  a  little 
pressure  on  the  pencil  or  pen  point. 

In  removing  the  pencil  or  pen  point  to  change  them,  be  very 
careful  to  pull  them  out  straight ;  do  not  bend  them  from  side  to 
side,  in  order  to  get  them  out  more  easily,  as  it  would  enlarge 
the  socket  and  consequently  spoil  the  instrument  for  accurate 
work. 


INSTRUMENTS    AND   THEIll    USES.  7 

Id  drawing  a  circle  of  larger  radius  than  could  be  drawn  with 
the  compass  in  its  usual  form  a  lengthening  bar  is  used.  In  this 
case  steady  the  needle  point  with  one  hand  and  describe  the 
circle  with  the  other. 

The  large  compasses  are  too  heavy  and  clumsy  to  make  small 
circles  nicely,  hence  the  bow  compasses  should  be  used  in  mak- 
ing all  circles  smaller  than  three-quarters  of  au  inch  radius,  or 
thereabouts,  depending  on  the  stiffness  of  the  spring.  Be  very 
careful  to  adjust  the  needle  point  to  the  same  length  as  the  pen- 
cil or  pen  point.  In  changing  the  radius  of  the  bow  compasses 
or  spacers,  press  the  points  together,  thus  removing  the  pressure 
from  the  nuts,  before  turning  the  nuts  in  either  direction.  The 
screw  thread  will  last  much  longer  if  this  is  done. 

4.  Dividers  or  Spacers.  These  are  used  to  lay  off  distances 
from  scales,  or  from  other  parts  of  a  drawing  to  a  line,  or  to 
divide  a  line  into  equal  parts.  In  laying  off  the  same  distance 
several  times  on  a  line,  keep  one  of  the  points  of  the  dividers  on 
the  line  all  the  time,  and  turn  the  instrument  in  an  opposite 
direction  each  time,  so  that  the  moving  point  will  pass~alter- 
nately  to  the  right  and  left  of  the  line.  Do  not  make  holes  in 
the  paper  in  doing  this,  as  it  is  impossible  to  ink  nicely  over 
them ;  a  very  slight  puncture  is  sufficient. 

5.  T-Squa.re.  The  T-square  should  be  used  with  the  head 
against  the  left-hand  edge  of  the  drawing  board  (unless  the  per- 
son is  left-handed),  and  horizontal  lines  only  should  be  drawn 
with  it.  Lines  perpendicular  to  these  should  not  be  drawn  by 
using  the  head  of  the  T-square  against  an  adjoining  edge  of  the 
board,  as  there  is  nO  pains  taken  to  make  these  edges  at  right 
angles  to  each  other,  but  they  should  be  drawn  by  using  the 
triangle  in  connection  with  the  T-square. 


8 


INSTRUMENTS   AND   THEIR   USES. 


Lines  should  be  drawn  with  the  upper  edge  only  of  the 
T-square. 

In  case  you  wish  to  use  the  T-square  as  a  guide  for  the  knife  in 
cutting  paper  to  size,  do  not  use  the  upper  edge  as  a  guide,  but 
turn  the  T-square  over  and  use  the  bottom  edge.  For,  unless 
you  are  very  careful,  the  knife  will  nick  the  edge,  which  would 
render  it  unfit  to  draw  lines  with. 

6.  Triangles.  To  draw  lines  which  shall  be  parallel  to 
another  by  means  of  the  triangles.     Let  AB  be  the  given  line. 


Place  either  edge  of  either  triangle  so  as  to  coincide  exactly 
with  the  given  line.  Place  the  other  triangle  (or  any  straight 

edge)  against  one  of  the  other  edges  of  the  first  triangle.  Then, 
holding  the  second  triangle,  or  straight  edge,  securely  in  this 
position  with  the  left  hand,  move  the  first  one,  still  keeping  the 
two  edges  in  contact.  Any  line  drawn  along  the  edge  which 
originally  coincided  with  the  line  AB  will  be  parallel  to  it. 

To  draw  lines  which  shall  be  perpendicular  to  another  by  means 
of  the  triangles.  Let  AB  be  the  given  Fine.  Place  the  longest 
side  of  the  triangle  so  as  to  coincide  exactly  with  the  given  line. 


INSTRUMENTS   AND   THEIR    USES.  9 

Place  the  other  triangle  (or  any  straight  edge)  against  one  of 
the  other  edges  of  the  first  triangle.  Then,  holding  the  second 
triangle  securely  in  this  position  with  the  left  hand,  revolve  the 
first  one  so  that  its  third  edge  is  against  the  second  triangle  or 
straight  edge.  Any  line  drawn  along  the  edge  which  originally 
coincided  with  the  given  line  AB  will  be  perpendicular  to  that 
line. 

The  right-hand  portion  of  the  figure  shows  how  the  two  tri- 
angles may  be  used,  in  connection  with  the  T-square,  to  draw 
lines  making  angles  of  15°  and  75°  with  a  given  line  (in  this 
case  the  line  which  coincides  with  the  edge  of  the  T-square). 
By  turning  the  triangles  over,  these  angles  may  be  drawn  in  the 
opposite  direction. 

Lines  making  angles  of  30°,  45°,  and  60°  are  drawn  directly 
by  means  of  one  triangle  and  T-square  or  straight  edge. 

7.  Irregular  Curves.  To  trace  an  irregular  curve  through 
a  series  of  points,  use  that  part  of  the  edge  of  the  curve  which 
coincides  with  the  greatest  possible  number  of  points  (never  less 
than  three),  and  draw  the  curve  through  these  points,  then  shift 
the  curve  so  as  to  coincide  with  other  points  in  the  same  wav, 
letting  the  instrument  run  back  on  a  part  of  the  curve  already 
drawn,  so  that  a  continuous  smooth  curved  line  may  be  formed. 

It  requires  a  considerable  practice  to  draw  irregular  curves  by 
means  of  an  instrument,  the  tendency  being  to  make  a  series  of 
loops,  on  account  of  some  of  the  points  being  covered  up. 
There  is  no  better  way  to  put  in  a  curve  in  pencil  than  by  doing 
it  free-hand,  provided  the  hand  and  eye  have  been  properly 
trained.  Of  course  the  curve  cannot  be  inked  in  free-hand  ;  the 
irregular  curve  must  be  used,  but,  being  no  longer  confined  to 
points,  it  is  not  difficult. 


10  INSTRUMENTS    AND   THEIR   USES. 

8.  Scales.  As  it  is  frequently  impossible  to  make  a  draw- 
ing on  paper  the  real  size  of  the  object,  it  is  customary  to  reduce 
the  actual  measurements  by  means  of  an  instrument  called  a 
scale, —  that  is,  the  drawing  may  be  made  •£,  \,  •£,  Tl$,  etc.,  size, 
according  as  the  relative  size  of  the  object  and  drawing  may 
require. 

If  it  is  desired  to  make  a  drawing  £  size,  then  3  inches  on 
the  drawing  will  represent  one  foot  on  the  object.  It  is  fre- 
quently necessary  to  represent  inches  and  fractions  of  an  inch, 
hence  divide  the  3  inches  into  12  equal  parts,  and  each  of  these 
parts  will  represent  one  inch  on  the  object.  If  each  of  the  12 
parts  are  subdivided  into  2,  4,  or  8  parts,  each  part  would  re- 
present respectively  £,  £,  or  £  of  an  inch  on  the  object.  This 
may  be  designated,  scale,  3  inches  equal  one  foot,  or  £  size. 

Ou  the  scale,  one  inch  equal  one  foot,  the  unit,  one  inch,  is 
divided  into  12  parts  to  represent  inches  as  before.  Thus,  to  make 
a  scale  of  any  unit  to  one  foot,  it  is  simply  necessary  to  divide 
that  unit  into  12  parts  to  represent  inches,  subdividing  these 
parts,  as  far  as  possible,  to  represent  fractions  of  an  inch. 

If  the  smallest  division  on  a  scale  represents  £  of  an  inch  on 
the  object,  the  scale  is  said  to  read  to  ■£  of  an  inch. 

The  student  will  find  on  his  triangular  scale  ten  different 
scales,  viz.,  t£t,  •£,  T3^,  £,  §-,  ^,  f ,  1,  l£,  and  3  inches  to  the  foot, 
reading  to  3",  1",  1",  1",  1",  £",  f ,  £",  ±",  and  £",  respectively, 
(the  double  prime  over  a  number  or  fraction  means  inches,  the 
single  prime  indicates  feet). 

The  scale  should  never  be  used  as  a  ruler  to  draw  lines  with. 

9.  Needle  Point.  Each  student  should  procure  a  fine 
needle,  break  off  the  eye  end,  and  force  the  broken  end  into  a 
.small,  round  piece  of  soft  pine  wood.  This  is  to  be  used  in 
pricking  off  measurements  from  the  scale,  marking  the  exact 


INSTRUMENTS   AND   THEIR   USES, 


H 


Cap,    .     .     .     . 

13  x  17  inches. 

Demy,      .     . 

15  x  20       " 

Medium, .     . 

17x22       " 

Royal,      .     . 

19x24       " 

Super-Royal, 

.     19  x  27       " 

Imperial, 

22x30       " 

intersection  of  two  lines,  etc.  Here,  as  in  the  case  of  the  needle 
point  in  the  compasses,  it  should  not  be  forced  into  the  paper : 
the  finest  puncture  possible  is  sufficient. 

10.  Drawing  Paper.  This  paper  comes  in  sheets  of 
standard  sizes,  as  follows:  — 

Elephant,  .  .  23  x  28  inches. 
Columbia,  .  .  23  x  34 
Atlas,  .  .  .  26  x  34 
Double  Elephant,27  x  40 
Antiquarian,  .  31  x  53 
Emperor,      .     .     48  x  68 

Whatman's  paper  is  considered  the  best.  This  paper  is  either 
hot  or  cold  pressed,  the  hot  pressed  being  smooth  and  the  cold 
pressed  rough.  The  rough  paper  is  better  for  tinting  work,  the 
smooth  takes  ink  lines  better  than  the  rough,  but  erasures  show 
much  more  distinctly  on  it,  hence  the  cold  pressed  is  better  for 
general  work.  The  names  of  the  sizes  of  the  paper  given  above 
have  no  reference  to  quality.  There  is  very  little  difference 
in  the  two  sides  of  the  paper,  but  that  one  which  shows  the 
maker's  name  in  water  lines,  when  held  up  to  the  light,  is  con- 
sidered the  right  side. 

11.  Thumb  Tacks.  These  are  used  for  fastening  the  paper 
to  the  drawing  board  when  it  is  not  necessary  to  stretch  it. 

12.  The  geometrical  problems  in  the  next  chapter  are  not 
given  with  the  view  of  teaching  geometry,  but  to  give  the  stu- 
dent practice  in  the  accurate  use  of  his  instruments. 

In  order  that  the  degree  of  accuracy  of  the  execution  of  the 
problems  may  be  readily  seen,  these  problems  will  not  be  inked. 

13.  The  plates  on  which  these  problems  are  to  be  drawn 
should  be  laid  out  10''  by  14"  (and  cut  this  size  when  finished) 


12  INSTRUMENTS   AND  THEIR   USES. 

with  a  border  line  one  inch  from  each  edge.  That  portion  of 
the  plate  within  the  border  line  is  to  be  divided  into  6  equal 
squares,  in  each  of  which  one  problem  is  to  be  drawn,  beginning 
with  No.  1  in  the  upper  left-hand  corner  square,  No.  2  in  the 
upper  middle,  No.  3  in  the  upper  right,  etc. 

The  number  of  the  plate  is  to  be  printed  in  the  upper  right- 
hand  corner  of  the  plate,  about  one-eighth  of  an  inch  above  the 
border  line,  and  the  student's  name  in  the  lower  right-hand  cor- 
ner, about  one-eighth  of  an  inch  below  the  border  line. 

Prou.  1,  11,  etc.,  as  the  case  may  be,  is  to  be  printed  in  the 
upper  right-hand  corner  of  each  square.  The  initial  letters  of 
the  name  should  be  made  T3^  of  an  inch  high,  and  the  small 
letters  ^  of  an  inch  high. 

The  letters  are  to  be  made  like  the  samples  furnished  in  the 
drawing  room,  and  they  should  be  made  as  nicely  as  possible. 


CHAPTER   II. 


GEOMETRICAL    PROBLEMS. 


Prob.  1,    To  bisect  a  straight  line  AB,  or  arc  of  a  circle  AFB. 
With  A  and  B  as  centres    and  any  radius 
greater  than  one  half  AB  draw  arcs  intersect- 
ing   in  C  and  D.      Join  CD.      CD  is  perpen- 
B  dicular  to  AB,  and   E  and  F  are  the  middle 
\i/  points  required. 

Note.  To  draw  a  perpendicular  to  a  line 
at  any  point  in  it,  as  E  in  AB.  Lay  off  equal  distances  EA 
and  EB  on  each  side  of  E,  and  proceed  as  above. 


Prob.  2.     From  a  point  C  outside  a  straight  line  AB  to  draw 
a  perpendicular  to  the  line. 

Wjth  C  as  a  centre    and  any  convenient 

radius  cut  AB  in  the  points  A  and  B.       With 

B  ,  A  and  B  as  centres  and  any  radius  draw  arcs 

_„-'      intersecting  in  D.     Join  C  and  D,  and  CD  is 

vi/  the  perpendicular  required. 


14  GEOMETRICAL    PROBLEMS. 

Prob.  3.    To  draw  a  perpendicular  to  a  line  ATS  from  a  point 
C  nearly  or  quite  over  its  end. 

Draw  a  line  from  C  to  meet  AB  in  any 
point  B.  Bisect  BC  in  D  by  Prob.  1.  With 
1)  as  a  centre  and  radius  DC  draw  the  arc 
CAB  meeting  AB  in  A.  Draw  through  A 
and  C. 


.    r^- 

'*    '-----■-''  Note.     If  a  perpendicular  be  required  at 

A,  take  any  point  D  as  a  centre  and  radius  DA  and  draw  an 
arc  CAB.  Through  B  and  D  draw  a  line  to  meet  the  arc  in  C. 
Draw  through  C  and  A. 


Prob.  4.    To  draw  a  perpendicular  to  a  line  AB  from  a  point 
A  at  or  near  its  end. 

With  A  as  a  centre   and  any  radius   draw 

the  arc  CD.     With  centre  D  and  same  radius 

cut  CD  in  C.     With  C  as  a  centre  and  same 

\\  radius    draw  an  arc  over  A,  and  draw  a  line 

\\  through  D  and  C,  producing  it  to  meet  this 

arc  in  E.     Draw  through  A  and  E. 


/T 


Prob.  5.     A  second  method. 

With  centre  A  and  any  radius  draw  an  arc 

CDE  ;  with  centre  C  and  same  radius  cut  this 

arc  in  D  ;  with  centre  D  and  same  radius  draw 

\  arc  EF ;  with  centre  E  and  same  radius  draw 

1     g    arc  intersecting  EF  in  F.     Join  AF. 


GEOMETRICAL   PROBLEMS. 


15 


Prob.  6.  To  construct  an  angle  equal  to  a  given  angle  CAB. 
Draw  any  line  as  DE,  and  take  any  point  in 
it  as  D.  With  A  as  a  centre  and  any  radius 
cut  the  sides  of  the  given  angle  in  B  and  C. 
With  centre  D  and  same  radius  draw  arc  EF. 
With  BC  as  a  radius  and  E  as  a  centre  cut  arc 
EF  in  F.     Join  DF. 

Note.     For  accuracy  of  construction  in  drawing  this  prob- 
lem the  longer  the  radius  AB  is  the  better. 

Prob.  7.      Through  a  given  point  C  to  draw  a  line  parallel  to 
a  given  line  AB. 

o^= c_  From    C    as    a    centre    and    any   radius 

\  s'     \    draw  the  arc  AD  ;  and  from  A  as  a  centre, 

\        ,''  \   with  the  same   radius,  draw  the   arc  BC. 

— ■- Si.  With  BC  as  a  radius,  and  A  as  a  centre, 

draw  an  arc  cutting  the  arc  AD  in  D.     Draw  through  D  and  C. 

Prob.  8.      To  draw  a  line  parallel  to  a  given  line  AB  at  a 
given  distance  CD  from  it. 

From  any  two  points  A  and  B  on  the  line 
as  centres  and  with  CD  as  a  radius  draw  arcs 
E  and  F.  At  A  and  B  erect  perpendiculars 
to  meet  the  arcs  in  E  and  F.     Draw  a  line 


c^ 


through  E  and  F. 


Prob.  9.      To  bisect  a  given  angle  BAC. 

With   A   as  a  centre   and    with    any    radius 

draw  the  are  BC,  cutting  the  sides  of  the  angle 

''   in  B  and  C.     With  centres   B  and  C  and  any 

radius  draw  arcs  intersecting  in  D.    Draw  AD. 


16 


GEOMETRICAL    PROBLEMS. 


Prob.  1 0.      To  divide  a  given  line  AB  into  any  number  of 
equal  parts.     (In  this  case  6.) 

,  f       From  A  draw  an  indefinite  line  A,  1,  2  •  •  5 

?,  ?'/  1\      at  any  angle  with  AB.     At  B  draw  Ba  •  •  •  e, 

A,  -,'   ,'   /  /   /  1b    making   the   angle  ABe  equal   to  the  angle 

',/  /   /   ;-  a  BA5.     With  any  distance  as  a  unit,  lav  off  on 

i,    jib  •'  J 

;\!d    b  the  lines  from  A  and  B  as  many  equal  spaces 

as   the   number  of  parts   required  less   one. 
Join  ]r,  '2d,  ?>c,  etc. 


Prob.  11.     Another  method. 
(say)  five  parts. 

■i 

9-' 


To   divide    a  line   AB   into 


Draw  A  1  •  •  •  •  5  at  any  angle  to  AB,  and 
lay  off  on  it  five  equal  spaces,  using  any  con- 
venient unit.        Join   5B.    and    through    the 
points    1,  2,  o,  4  draw  lines  parallel  to  Bo, 
meeting  AB  in  points  a,  b,  c,  d. 


Prob.  12.      To  divide  a  line  AC  into  the  same  proportional 
parts  as  a  given  divided  line  AB. 

Draw  from  a  point  A  the  lines  AC  and 
AB,  making  any  angle.  Join  B  and  C. 
Through  the  points  1,  2,  and  3  on  AB  draw 
lines  parallel  to  BC,  meeting  AC  in  points  a, 
b,  and  c,  the  required  points  of  division. 


GEOMETRICAL    PROBLEMS. 


17 


Prob.  13. 


C'a.  \b. 


Second  method. 

Let  M  be  the  line  to  be  divided  into  parts 
proportional  to  the  parts  Al,  12,  23,  and  3B 
of  the  line  AB.      Draw  M  parallel   to  AB  at 
CD  by  Prob.  8.     Draw  lines  through  AC  and 
N  BD  to  meet  in  E.     Through  E  draw  lines  to 
^ 1,  2,  and  3,  cutting  CD  in  a,  b,  and  c,  the  re- 
quired points  of  division. 

Note.     If  the  parts  on  AB  are  equal,  the  parts  on  CD  will 
be  equal. 


--D 


-r 


Prob.  14.  To  draw  on  a  given  line  AB  as  an  hypothenuse  a 
right  triangle  with  its  sides  having  the  proportion  of  3,  4,  and  5. 
Divide  AB  by  Prob.  11  into  5  equal  parts. 
With  centre  B  and  a  radius  equal  to  three  of 
the  parts  draw  an  arc,  and  with  centre  A  and 
a  radius  equal  to  four  of  the  parts  cut  this 
arc  in  C.     Join  AC  and  BC. 


Prob.  15. 


To  trisect  a  right  angle  CAB. 

With  centre  A  and  any  radius  draw  the  arc 
of  the  quadrant,  cutting  the  sides  in  C  and  B. 
With  centres  C  and  B  and  the  same  radius  cut 
the  arc  in  points  1  and  2.     Join  Al  and  A2. 

Note.      The  angle  1AB  is  an  angle  of  60°, 


the  construction  of  which  is  apparent. 


18 


GEOMETRICAL   PROHLKMK. 


Prob.  16. 

and  N. 


C Bi 


To  find  a  mean  -proportional  between  two  lines  M 

Upon  an  indefinite  line  AB  lay  off  AC  equal 
to  M,  and  CB  equal  to  N.  Bisect  AB  in  E 
by  Prob.  1,  and  with  centre  E  and  radius  EB 
draw  a  semicircle.  At  C  draw  CD  perpen- 
dicular to  AB  (Prob.  5).  CD  is  the  mean  pro- 
portional required. 


Prob.  17. 
and  P. 


To  find  a  fourth  proportional  to  three  lines  M,  N, 


Draw  AB  equal  to  M,  and  at  any  conve- 
nient angle  draw  AC  equal  to  N.  Upon 
AB  produced  make  BD  equal  to  P.  Join 
BC,  and  draw  DE  parallel  to  BC,  to  meet 
AC  produced  in  E.     CE  is  the  fourth  pro- 


,b__jL 


portional  required. 


Prob.  18.     At  a  point  A  in  a  line  AC  to  make  an  angle  of 
30°. 

With  any  point  B  as  a  centre  and  radius 
AB  draw  a  semicircle  ADC.     With  centre  C 
and  same  radius  cut   this  arc  in   D.     Draw 
-  AD.     DAC  =  30°. 


Prob.  19.     Having  given  the  sides  AB,  M,  and  N  of  a  tri- 
angle to  construct  the  figure. 

~^c/  With  centre  A  and  radius  M  draw  an  arc. 

With  B  as  a  centre  and  radius  N  draw  an  arc 
to  cut  the  first  arc  at  C.     Join  AC  and  BC. 


GEOMETRICAL   PROBLEMS. 


19 


Prob.  20.      On  a  given  side  AB  to  construct  a  square. 
1Q 

Draw  BD  at  right  angles  to  AB  and  equal 
)    to  AB  (Prob.  5).     With  A  and  D  as  centres 
and  radius  AB  draw  arcs  intersecting  in  C. 
Join  AC  and  CD. 


Prob. 
and  M. 
t 


21.       To    construct  a  rectangle    of  given    sides    AB 


At  B  draw  BD  perpendicular  to  AB  by 
Prob.  5,  and  equal  to  M.  With  A  as  a  cen- 
tre and  radius  equal  to  M  draw  an  arc,  and 
from  D  as  a  centre  and  a  radius  equal  to  AB 
cut  this  arc  in  C.     Join  AC  and  CD. 


Prob.  22.      On  a  given  diagonal  AB  to'  construct  a  rhombus 
of  given  side  AC. 

With  centres  A  and  B  and  radius  AC  draw 
arcs  intersecting  in   C  and  D.     Join  AC,  AD, 
6  BC,  and  BD. 


Prob.  23. 


On  a  given  base  AB  to  construct  a  pentagon. 

With  centres  A  and  B  and  radius  AB 
draw  circles  intersecting  in  1  and  2.  Join 
1  and  2.  With  centre  2  and  same  radius 
draw  the  circle  3A5B4,  giving  points 
3  and  4.  Produce  35  to  C  and  45  to 
E.  With  centres  C  and  E  and  radius 
AB  draw  arcs  intersecting  in  D.  Draw 
BCDEA. 
Note.   This  is  an  approximate  method. 


20 


GEOMETRICAL    PROBLEMS. 


Prob.  24.  To  construct  a  regular  hexagon  of  given  side  AB. 
With  A  and  B  as  centres  and  radius  AB  draw 
arcs  intersecting  in  0.  With  centre  O  and  radius 
AB  draw  a  circle,  and  lay  off  BC,  CD,  etc., 
each  equal  to  AB.  Join  the  points  B,  C,  D, 
E,  F,  and  A. 

Note.     The  radius  of  any  circle  goes  around  the  circumfer- 
ence as  a  chord  six  times  exactly. 

Prob.  25.      On  a  given  base  AB  to  construct  a  regular  poly- 
gon of  any  number  of  sides  (in  this  case  7). 

With    centre   A    and   radius  AB  draw  a 

semicircle  and  divide  it  at  points  1,  2,  3,  4, 

\|  etc.,  into  as  many  equal   parts  as  there  are 

sides  in  the  required  polygon.      Draw  a  line 

from  the  second  point  of  division  2  to  A. 

2 A  is  one  side  of  the  required  polygon. 

Bisect   AB   and   A2   hy  perpendiculars,  by 

Prob.  1,  meeting  in  D.     With  D  as  a  centre  and  radius  DA 

draw  the  circle  BA2,  etc.     Apply  AB  as  a  chord  to  the  circle  as 

many  times  as  there  are  sides  in  the  j^olygou. 


J-^---- 


rF^B 


Prob.  26.      On  a  given  line  AB  to  construct  a  polygon  of  any 
number  of  sides.     An  approximate  method. 

Bisect  AB,  and  produce  the  bisecting 
line  indefinitely.  With  centre  A  and 
radius  AB  draw  the  arc  BC,  cutting  the 
bisecting  line  in  C.  Divide  the  arc  BC 
into  six  equal  parts,  in  points  1,  2.  3,  etc. 

To  construct  a  pentagon.  With  cen- 
tre C  and  radius  CI  draw  an  arc,  cutting 
the   bisecting  line  in  a  point  below   C, 


GEOMETRICAL   PROBLEMS. 


21 


which  is  the  centre  of  the  circle  circumscribing  the  pentagon.  If 
the  polygon  is  to  have  more  sides  than  six,  set  up  from  C  on  the 
line  Cab  as  many  parts  of  the  arc  CB  as  added  to  six  make 
the  number  of  sides  of  the  required  polygon  ;  thus,  for  a  seven- 
sided  polygon  set  up  one  division  as  Ca ;  for  an  eight-sided  set 
up  two  divisions  as  Cab,  and  so  on.  a,  b,  c,  d,  etc.,  are  the  cen- 
tres for  the  circumscribing  circles  of  the  polygons,  each  side  of 
which  is  equal  to  AB. 


PrOB.  27.  On  a  given  base  AB  to  construct  a  regular  octa- 
gon. 

At  A  and  B  erect  perpendiculars  by 
Prob.  5  to  AB,  and  bisect  the  exterior  right 
angles,  making  the  bisectors  AC  and  BD 
each  equal  to  AB.  Draw  CD,  cutting  the 
perpendiculars  in  E  and  F.  Lay  off  EF 
a  b""  "'  from  E  to  G  and  from  F  to  H.  Draw 
an  indefinite  line  through  GH.  Make  GK,  GL.  HN  and  IIM 
each  equal  to  CE  or  FD.      Connect  C.  K,  L,  M,  N,  and  D. 

Note.  This  may  be  done  by  Prob.  25.  Check  the  con- 
struction by  seeing  if  AN,  CM,  BD,  and  KL  are  parallel. 
CA,  KB,  etc.  should  be  parallel ;  so  also  AD,  CN,  and  KM. 


Prob.  28.      To  construct  an  octagon  within  a  square  ABCD. 

Draw  the  diagonals  AC  and  BD.  With  A, 
B,  C,  and  D  as  centres  and  the  half  of  the 
diagonal  of  the  square  as  a  radius  draw  the 
arcs  EF,  GH,  KL,  and  MN.  Join  the  points 
MG,  FL,  HN,  and  KE. 


D 

c 

^~~-%       L 

',-'    H 

X 

B.---5/'        » 

V-  -  -  L 

i 

22 


GEOMETRICAL   PROBLEMS. 


Prob.  29.      The  altitude  AB  of  an  equilateral  triangle  being 
given,  to  construct  the  triangle. 

Draw  CD  and  EF,  both  perpendicular  to 
^  AB.  With  A  as  a  centre  and  any  radius 
describe  the  semicircle  CGHD.  With  C  and 
D  as  centres  and  the  same  radius  draw  arcs 
cutting  the  semicircle  in  G  and  H.  Draw 
AGE  and  AHF. 

Prob.  30.      Given  the  base  AB  of  an  isosceles  triangle,  and 
the  angle  at  the  vertex  M,  to  draw  the  triangle. 

With  centre  B  and  any  radius  draw  a 
semicircle  cutting  the  base  produced  at  C. 
Make  the  angle  DBC  equal  to  M  by  Prob. 
6.  Bisect  ABD  by  Prob.  9.  Make  the 
angle  FAB  equal  to  EBA,  and  produce 
AF  and  BE  to  meet  in  G. 


Prob. 
a  square 


31.      Within  a  regidar  hexagon  ABCDEF  to  inscribe 

St- 


Draw  a  diagonal  AD.  Bisect  AD  by 
a  perpendicular  213  (Prob.  1).  Bisect  by 
Prob.  9  the  right  angles  21 A  and  2 IB,  and 
produce  the  bisectors  to  meet  the  sides  of  the 
hexagon  in  points  G,  H,  L,  and  K.  Join  G, 
H,  L,  and  K. 


Prob.  32.      On  a  given  diagonal  AB  to  construct  a  square. 


Bisect  AB  in  O  (Prob.  1)  ;  and  with  centre 
0  and  radius  OA  draw  a  circle  to  cut  the  bisect- 
ing line  in  C  and  D.  Draw  AC,  CB,  BD,  and 
DA. 


GEOMETRICAL   PROBLEMS. 


23 


Prob.  33.     Within  a  given  triangle  ABC  to  inscribe  a  square. 


Draw  AD  perpendicular  to  and  equal  to  AB. 
From  C  draw  CE  perpendicular  to  AB  (Prob.  2). 
Draw  DE  cutting  AC  in  F.  From  F  draw  FK 
perpendicular  to  AB.  Make  KH  equal  to  KF, 
and  from  H  with  radius  HK  cut  BC  in  G.     Join 


EH    B 


FG  and  GH. 


Prob.  34.     About  a  given  circle  FEGD  to  circumscribe  an 
equilateral  triangle. 

Draw  the  diameter  DE.  With  centre  E 
and  radius  of  the  given  circle  draw  the  cir- 
cle AFG.  Prolong  DE  to  A.  With  centres 
D,  F,  and  G  and  radius  DG  draw  arcs  inter- 
secting at  B  and  C.     Join  AB  and  AC. 


To  circumscribe  a  circle  about  a  triangle  ABC. 

By  Prob.  1   bisect  two  of  the   sides  AB 
^j.  and   BC  by   perpendiculars   meeting   in  0. 
With   centre    O   and   radius  OA  draw  the 
circle. 

Note  1.     If  any  three  points  are  given 
not   in   the    same  straight  line,   a  circle  is 
passed  through   them  by  joining  the  points 
and  proceeding  as  above. 

Note  2.    If  any  circle  is  given,  its  centre  is  found  by  taking 
any  three  points  in  its  circumference,  and  proceeding  as  above. 


■24 


GEOMETRICAL    PROBLEMS. 


Puob.  36.      To  inscribe  a  pentagon  within  a  circle. 

Draw  any  diameter  AB,  and  a  radius  CD 
perpendicular  to  it.  Bisect  BC  in  E.  With 
centre  E  and  radius  ED  draw  the  arc  DE. 
B  With  centre  D  and  radius  DF  draw  the  arc 
FG.  DG  is  the  side  of  the  required  pen- 
tagon. 

To  inscribe  a  hexagon  within  a  circle. 

Draw  the  diameter  AB,  and  with  centres  A 
and  B  and  the  radius  of  the  given  circle 
draw  arcs  COD  and  EOF,  cutting  the  circum- 
ference in  C,  D,  E,  and  F.  Join  AD,  DF, 
FB,  etc. 

Note.  Joining  points  A,  E,  and  F  gives 
an  inscribed  equilateral  triangle. 


Prob.  38.      To  construct  a  regular  polygon  of  any  number  of 
sides,  the  circumscribing  circle  being  given. 

Draw  any  diameter  AB  and  divide  it 
into  as  many  equal  parts  as  there  are  sides 
in  the  required  polygon  (in  this  case  5). 
With  A  and  B  as  centres  and  radius  AB 
draw  arcs  intersecting  in  C.  Draw  a  line 
from  C  through  the  second  point  of  division 
of  AB  to  meet  the  circumference  in  D. 
AD  is  one  side  of  the  required  polygon. 
Lay  off  AD  as  a  chord  as  many  times  as 
there  are  sides  in  the  required  polygon  (in 
this  case  o). 

Note.     This  is  an  approximate  method. 


;£' 


GEOMETRICAL    PROBLEMS. 


25 


Prob.  39.      To  circumscribe  a  hexagon  about  a  circle. 

Draw  the  diameters  AB  and  CD  perpen- 
dicular to  each  other.  Divide  each  quadrant 
into  thirds,  by  Prob.  15,  at  points  E,  F,  etc. 
Join  B  and  E,  cutting  CD  in  G.  With  G 
as  a  centre  and  radius  GE  draw  arc  ED, 
cutting  COD  in  D.  With  centre  O  and 
radius  OD  cut  the  diameters  produced  in  points  H,  K,  C,  L,  and 
M.     Join  points  H,  D,  M,  L,  C,  and  K. 

Note.  Practically  the  60°  triangle  placed  on  a  T-square 
whose  blade  is  parallel  to  COD  will  give  HD,  DM,  etc.,  by 
making  it  tangent  to  the  circle  at  E,  F,  etc.  KH  and  LM  are 
drawn  by  the  T-square. 


Prob.  40. 

a  circle. 


Prob.  41. 
without  it. 


To  circumscribe  a  square,  also  an  octagon,  about 

Draw  the  diameters  AB  and  CD  at  right 
angles  to  each  other.  With  centres  A,  B,  C, 
and  D  and  radius  OA  describe  arcs  intersect- 
ing in  points  E,  F,  G,  and  H.  These  points  con- 
nected give  a  square  about  the  circle.  Inscribe 
an  octagon  in  the  square  by  Prob.  28. 

To  draw  a  tangent  to  a  circle  B  from,  a  point  A 


Draw  AB  and  bisect  it  in  D.     With 

centre  D  and  radius  DA  draw  a  semicircle 

cutting  the  given  circle  in  C  and  E.    Join 

AC.     By  joining  AE  a  second  tangent 

is  found,  equal  to  AC. 

Notk.     To  draw  a  tangent  to  a  circle  from  a  point  C  on  the 

circumference.     Join  BC,  and  at  C  draw  AC  perpendicular  to 

BC.     For  the  tangent  is  perpendicular  to  the  radius  at  the  jioiuf 

of  tangency. 


26 


GEOMETRICAL    PROBLEMS. 


Prop..  42.      To  draw  a  tangent  to  the  arc  of  a  circle  when 
the  centre  is  not  accessible. 

Let  C  be  the  point  upon  the  given  arc, 
AB,  at  which  the  tangent  is  to  be  drawn. 
Lay  off  equal  distances  upon  the  arc  from 
C  to  A  and  B.  Join  A  and  B.  Through  C  draw  a  line  par- 
allel to  AB  by  Prob.  7. 

Prob.  43.      To  draw  a  tangent  at  a  given  point  A  on  a  circle 
when  the  "preceding  method  is  not  applicable. 

From  A  draw  any  chord  AB.  Bisect  AB  in 
C,  and  the  arc  ADB  in  D  by  Prob.  1.  With 
A  as  a  centre  and  a  radius  AD  draw  the  arc 
EF.  With  D  as  a  centre  and  radius  DF  draw 
an  arc  cutting  EF  at  E.     Join  AE. 

Prob.  44.  To  draw  a  tangent  to  two  given  circles,  A  and  B. 
Through  A  and  B  draw  a  line. 
Make  DH  equal  to  the  radius  BF. 
Draw  the  circle  A-HC,  and  from  B 
draw  the  tangent  BC  by  Prob.  41. 
Draw  AC  and  produce  it  to  E.  Make 
the  angle  FBK  equal  to  CATI.     Join  EF. 


Prob.  45.     To  draw  a  tangent  to  two  given  circles  which  shall 
pass  between  them. 

Join  A  and  B,  and  draw  AD  and  BC 
perpendicular  to  AB.  Draw  DC,  cut- 
ting AB  in  E.  Draw  a  tangent  from 
E  by  Prob.  41  to  the  given  circles. 
Join  the  tangeut  points  F  and  G.  FG 
is  the  required  tangent. 


GEOMETRICAL   PROBLEMS. 


27 


Prob.  46.  To  draw  a  circle  tangent  to  a  given  line  AB  at 
a  given  point  B  in  it,  which  shall  also  pass  through  a  fixed  paint 
C  without  the  line. 

Draw  BD  perpendicular  to  AB,  at  the 
point  B.  Join  CB,  and  draw  a  perpen- 
dicular to  it  at  its  middle  point,  Prob.  1. 
The  intersection  of  this  perpendicular  and 
BD  gives  D  the  centre  required. 


Prob.  47.  To  draw  a  circle  of  a  given  radius  AB,  which 
shall  be  tangent  to  a  given  circle  C,  and  also  to  a  straight  line 
DE. 

Draw  GH  parallel  to  DE,  by  Prob. 
8,  at  the  distance  AB.     With  a  radius 
CM,  equal   to    the   radius  of  circle  C 
plus  AB,  draw  an  arc  to  meet  GH  in  0. 
With  the  centre  O  and  radius  AB  draw 
the  required  circle. 
Note.     If  two  circles  are  tangent,  the  straight  line  connect- 
ing the  centres  passes   through  the  point  of  tangency.     This 
point  it  is  very  important  to  locate  precisely  in  all  cases  of  tan- 
gency. 

Prob.  48.  To  draw  a  circle  tangent  to  a  given  circle  D,  and 
also  tangent  to  agiven  line  AB,  at  a  given  point  B  on  the  line. 

Draw  BC  perpendicular  to  AB,  and 
make  BC  equal  to  the  radius  of  D. 
Join  DC,  and  at  its  middle  point  draw 
a  perpendicular  to  meet  BC  in  E,  the 
required  centre. 

Note.  By  laying  off  BC  above  the 
line  AB,  and  proceeding  as  above,  another  circle  is  found  helow 
AB. 


28 


< ;  E(  tM  ETUICAL    PROBLEMS. 


Prop..  49.      To  draw  a  circle  tangent  to  a  given  circle  A,  and 
a  given  line  BC,  at  a  given  point  E  on  the  circle. 

Draw  AE  and  produce  it.  At  E 
draw  a  perpendicular  to  AE,  meeting 
BC  in  B.  Draw  the  bisector  of  the 
angle  EBC,  meeting  AE  produced 
in  D.     DE  is  the  radius  required! 

Note.    By  bisecting  EBM  another 
circle,  whose  centre  is  G,  is  found, 
enclosing  the  circle  AE. 


Prob.  50.        To   draw  a  circle  of  given   radius  R  tangent  to 
two  given  lines.  AB  (aid  CD. 

Draw  lines  parallel  to  AB,  and 
CD,  at  the  given  distance  R  from 
them  by  Prob.  8,  meeting  in  E, 
the  required  centre. 


Prob.  51.      To  draw  any  number  of  circles  tangent  to  eacli 
other,  and  also  to  two  given  lines  AB  and  AD. 

Bisect  BAD  by  AC,  Prob.  9.     Let 
one  of  the  circles  be  45ED,  drawn  by 

_\c taking   C   as   a    centre,   and  a  radius 

/  equal  to  the  perpendicular  CD  from 

C  to  AD.  At  E  draw  EF  perpendic- 
ular to  AC.  With  centre  F  and  ra- 
dius FE  draw  the  arc  DEG,  cutting  AD  in  G.  At  G  make 
GH  perpendicular  to  AC.  H  is  a  centre  required.  Repeat 
the  process. 


GEOMETRICAL    PROBLEMS. 


29 


Prob.  52.      To  draw  a  circle  through  a  given  point  D  and 
tangent  to  two  given  lines  AB  and  AC. 

Draw  AD.  Bisect  BAG  by  AE. 
From  any  point  K,  on  AE  as  a 
centre,  draw  a  circle  tangent  to  AB 
and  AC,  and  cutting  AD  in  H. 
Draw  HK.  At  D  draw  DE,  mak- 
ing the  angle  ADE  equal  to  AHK. 
E  is  the  required  centre. 


Prob.  53.      To  draw  a  circle  of  a  given  radius  R  tangent  to 
two  given  circles  A  and  B. 

Through  A  and  B  draw  indefinite 
Hues,  and  make  DE  and  EG  each 
equal  to  R.  With  A  and  B  as  centres, 
and  radii  AE  and  BG,  draw  arcs  cut- 
ting each  other  at  C,  the  required  cen- 
tre. 
R     ^  Notk.     These  arcs  will  intersect  in 

a  second  centre. 


Prob.  54.      To  draw  a  circle  through  a  point  C,  and  tangent 
to  a  given  circle  A,  at  a  point  B  in  its  circumference. 


Draw  AB   and   produce  it.     Join  BC 
and  bisect  it  by  a  perpendicular  meeting 
-?Si^  AB  produced  in  D,  the  required  centre. 


30 


GEOMETRICAL    PROBLEMS. 


Prob.  55.  Given  two  parallel  straight  lives  AB  and  CD,  to 
draw  ares  of  circles  tangent  to  them  at  B  and  C,  and  pasting 
through  E,  which,  is  anywhere  on  the  line  BC. 

jc e_       At  B   and  C   erect  perpendiculars. 

Bisect  BE  and  CE  by  perpendiculars 
(Prob.  1),  meeting  the  perpendiculars 
from  B  and  C  in  F  and  H,  the  re- 
quired centres.      Draw   the  arcs   BE 

This  is  called  a  reversed  curve. 


Prob.  56.      To  draw  a  circle  tangent  to  two  given  circles  A 
and  B  at  a  given  point  C  in  one  of  them. 

Draw  a  line  indefinitely  through 
A  and  C.  Make  CD  and  CE  each 
equal  to  the  radius  of  B.  Join  BE 
and  BD.  Bisect  BE  and  BD  by 
perpendiculars  meeting  AC  pro- 
duced in  F  and  G,  the  centres  of 
the  required  circles. 

Note.  The  tangent  points  should 
be  determined  accurately. 


Prob.  57. 


Same  as  Prob.  56.      A  second  method. 

Draw  a  line  through  A  and  C 
indefinitely.  Through  B  draw  DE 
parallel  to  CA,  cutting  B  in  D  and 
E.  Draw  CD  to  F  and  CE.  cut- 
ting B  in  H.  Through  B  and  H 
draw  BHK,  cutting  CA  in  K,  one 
centre  required.  Through  B  and  F 
draw  FBL,  cutting   CA    in    L,  an- 


other centre  required. 


GEOMETRICAL   PROBLEMS. 


31 


Prob.  58.   Same  as  Prob.  56.  Prob.  59.   Same  as  Prob.  56. 

Method  of  Prob.  57. 


Probs.  60  and  61.     Same  as  Prob.  56.     Method  of  Prob.  57. 


Prob.  62.      To  inscribe  a  circle  within  a  triangle  ABC. 

Bisect  two  of  the  angles  of  the  triangle, 

> ':  say  A  and    B,  by   Prob.  9.     The  bisectors 

meet  in  D,  the  centre  of  the  required  circle. 

A  perpendicular  from  D  to  either  side  is  the 

~*  required  radius. 


39 


<; KO METRICAL   PROBLEMS. 


Prob.  63.  Within  an  equilateral  triangle  ABC  to  inscribe 
three  equal  circles,  each  touching  two  others,  and  two  sides  of  the 
triangle. 

Draw  the  bisectors  of  the  angles  A.  B, 
and  C,  cutting  the  sides  in  D,  E,  and  F. 
With  centres  D,  E,  and  F,  and  radius  DF, 
draw  arcs  cutting  the  bisectors  in  H,  L, 
and  K,  the  required  centres. 


Pkob,  64.      Within  an  equilateral  triangle  to  draw  three  equal 
circles,  each  touching  two  others  and  one  side  of  the  triangle. 

Bisect  the  angles  A,  B,  and  C.  Bi- 
sect the  angle  DAB  by  AG.  G  is  the 
ceutre  of  one  of  the  required  circles. 
With  the  centre  O  of  the  triangle  as  a 
centre,  and  radius  OG,  draw  a  circle 
cutting  AD  in  H  and  BE  in  K.  With 
centres  G,  H,  and  K,  and  radius  GF, 


\-K 


draw  the  circles  G,  H,  and  K. 


Prob.  65.  Within  an  equilateral  triangle  ABC  to  draw  six 
equal  circles  which  shall  be  tangent  to  each  other  and  the  sides  of 
the  triangle. 

Inscribe  the  three  circles  K,  H,  and 
<  i  by  Prob.  64.  Draw  LGM  parallel 
to  AB,  MIIN  parallel  to  BC,  and 
LKN  parallel  to  AC.  The  points  L, 
M,  and  N  are  the  centres  of  the  other 
three  circles. 


GEOMETRICAL   PROBLEMS. 


33 


Pkob.   66.      Within  a  square  ABCD  to  draw  four  equal  cir- 
cles each  touching  two  others  and  two  sides  of  the  square. 

Draw  the  diagonals  AC  and  BD,  and 
the  diameters  EF  and  GH.  Draw  ELI, 
HF,  FG,  and  GE,  giving  the  centres 
M,  N,  O,  and  P.  The  radius  OR  is  found 
by  joining  OP. 


Prob.  G7.     Within  a  given  square  ABCD  to  draw  four  equal 
circles  each  touching  t>vo  others  and  one  side  of  the  square. 

Draw  the  diagonals  AC  and  BD,  and 
the  diameters  EF  and  GIT.  Bisect  the 
angle  OAB  by  AK,  cutting  EF  in  K. 
*  With  radius  OK  and  centre  O  draw  a  cir- 
cle cutting  the  diameters  in  the  points  L, 
M,  N,  aud  K,  the  required  centres. 


Prob.  68.  Within  a  square  ABCD  to  draw  four  equal  semi- 
circles, each  tangent  to  two  sides  of  the  square,  and  their  diame- 
ters forming  a  square. 

Draw  the  diagonals  and  diameters.  Bi- 
sect FC  aud  Bli  in  L  and  K.  Draw  LK, 
cutting  GH  in  M.  Set  off  SM  from  S 
on  the  diameters  to  N,  O,  and  P.  Join 
the  points  M,  N,  O,  P.  The  intersections 
of  these  lines  with  the  diagonals  give  the 
required  centres  1,  2,  .'3,  aud  4. 


34 


GEOMETRICAL   PROBLEMS. 


PrOB.  69.  Within  a  square  ABCD  to  draw  four  equal  semi- 
circles, each  touching  one  side  of  the  square  and  their  diameters 
forming  a  square. 

Draw  the  diagonals  and  diameters.   Draw 

EH,  HF,  FG,  and  GE,  giving  the  points 

K,  L,  M,  and  N.     The  lines  joining  these 

L  points  cut  the  diameters  in  points  1,  2,  3, 

and  4,  the  required  centres. 


Pkob.  70.      To  draw  within  a  given  circle  ABC  three  equal 
circles  tangent  to  each  other  and  the  given  circle. 

Divide  the  circle  into  six  equal  parts 
by  diameters  AE,  DC,  etc.  (Prob.  24). 
Produce  any  diameter,  as  AE  to  G,  mak- 
ing EG  equal  to  the  radius  of  the  given 
circle.  Join  CG.  Bisect  the  angle  OGC 
-5  by  GH,  intersecting  OC  in  II.  With 
centre  O  and  radius  OH  draw  the  circle  HKLM.  K,  L.  and 
M  are  the  centres  of  the  circles. 


Puob.  71.      To  dram  within  a  given  circle  ACDB  four  equal 
circles  which  shall  be  tangent  to  each  other  and  the  given  circle. 

Draw  the  diameters  AB  and  CD  at  right 
angles,  and  complete  the  square  FBED. 
Draw  the  diagonal  FE  and  bisect  the  angle 
FED  by  EG;  FD  and  EG  meet  in  G. 
With  centre  F  and  radius  FG  draw  a  cir- 
cle cutting  the  diameters  in  G,  H,  K,  and  L, 
L^  the  required  centres. 


GEOMETRICAL   PROBLEMS. 


35 


Prob.  72.      Within  a  given   circle  AFD .  .  . .  C  to  draw  six 
equal  circles  tangent  to  each  other  and  the  given  circles. 

Draw  the  diameters  AB,  CD,  and  EF,  di- 
viding the  circle  into  six  equal  parts  (Prob. 
24,  Note).  Divide  any  radius  as  OB  into 
three  equal  parts  (Prob.  11),  at  points  F  and 
G.  With  centre  O  and  radius  OF  draw  a  cir- 
cle giving  the  centres  F,  1,2, 3,  4,  5  required. 
A  circle  of  the  radius  FB  may  be  drawn 
S  b  T  from  centre  O  tangent  to  the  six  circles. 
Note.  The  above  is  a  special  method.  The  general  method, 
to  draw  any  number  of  equal  circles  in  a  given  circle,  tangent 
to  each  other  and  the  given  circle,  is  to  divide  the  circle  by 
diameters  into  twice  as  many  equal  parts  as  circles  required. 
From  B,  the  extremity  of  any  one  of  these  diameters,  draw 
a  tangent  ST,  Prob.  41,  Note.  Produce  the  diameters  on 
each  side  of  AB  to  meet  the  tangent  in  S  and  T.  Bisect  the 
angle  T.  The  bisector  meets  OB  in  F,  the  centre  of  one  of  the 
required  circles.  Or  F  may  be  obtained  by  making  TM  equal 
to  TB,  and  at  M  drawing  a  perpendicular  to  OT,  meeting  OB 
in  F.  With  centre  O  and  radius  OF  draw  a  circle  cutting  the 
diameters  in  points  1,  2,  3,  4,  etc.,  the  centres  required. 

Prob.  73.     About  a.  given  circle  to  circumscribe  any  number 
of  equal  circles  tangent  to  each  other  and  the  given  circle. 

^ — i — ~„  Divide   the   circumference  of  the  (riven 

circle  by  diameters  into  twice  as  many  parts 
as  circles  required.  From  the  extremity  B 
of  any  diameter  draw  a  tangent  SBT  (Prob. 
41,  Note),  and  produce  the  diameters  on 
each  side  of  OB  to  meet  SBT  in  S  and  T. 
Produce  OT,  making  TN  equal  to  TB. 
Make  NR  perpendicular  to  TN,  meeting 


36 


GEOMETRICAL    PROBLEMS. 


OB  produced  in  R,  the  centre  of  one  of  the  required  circles. 
The  other  centres  are  at  the  intersection  of  a  circumference 
drawn  with  radius  OR  and  centre  0,  and  every  other  diameter 
produced. 


Prob.  74.  Within  a  given  circle  AC  .  . .  E  to  draw  any  num- 
ber of  equal  semicircles  touching  the  given  circle,  and  their  diam- 
eters forming  a  regular  polygon. 

In  the  given  circle  let  OA  and  OB 
be  two  radii  at  right  angles.  Divide 
the  given  circumference,  commencing 
at  B,  into  twice  as  many  parts  as  semi- 
circles required,  and  draw  diameters  to 
the  points  of  division.  Join  BA.  BA 
cuts  the  first  diameter  to  one  side  of  OB 
at  G.  G  is  one  end  of  two  adjacent 
diameters  required.  Lay  off  OG  from  O  on  every  other  diam- 
eter in  points  H,  K,  etc.  Join  UK,  KG,  etc.  1,  2,  3,  etc.,  are 
the  centres  of  the  required  semicircles. 


Prob.  75.      To  divide  a  circle  into  any  number  of  parts  which 
shall  be  equal  in  area. 

Let  the  number  of  parts  be  four.  Divide 
a  diameter  into  twice  as  many  equal  parts 
as  areas  required,  in  this  case  eight,  by 
points  1,  2,  3,  4,  etc.  With  1  and  7  as  cen- 
tres and  radius  01  describe  a  semicircle  on 
opposite  sides  of  the  diameter ;  with  cen- 
tres 2  and  6  and  radius  02  do  the  same 
thing,  and  so  continue  taking  each  point  as  a  centre  and  the  dis- 
tance from  it  to  the  end  of  the  diameter  as  a  radius. 


GEOMETRICAL   PROBLEMS. 


Prob.  7G.      To  divide  a  circle  into  concentric  rings  having 

equal  areas. 

Divide  the  radius  CD  into  as  many 
equal  parts  as  areas  required  (Prob.  11)  in 
points  1,  2,  3,  etc.  On  CD  as  a  diameter 
draw  a  semicircle,  and  at  the  points  1,  2, 
3,  etc.  draw  lines  perpendicular  to  CD, 
meeting  the  semicircle  in  points  A,  B,  etc. 
With  centre   C   aud   radii   CA,   CB,  etc. 

draw  concentric  circles. 


Prob.  77.  A  chord  AB  and  a  point  C  being  given,  to  find 
other  points  in  the  arc  of  the  eircle  passing  through  A.  B,  and  C 
without  using  the  centre. 

Draw  AB,  AC,  and  BC.  Suppose 
four  more  points  are  required.  With 
any  radius  and  centres  A  and  B  draw 
the  arcs  DE  and  FG.  Make  CAE 
equal  CBA,  and  GBC  equal  CAD. 
Divide  the  arcs  DE  and  FG  into  the 
same  number  of  equal  parts,  one 
more  than  the  number  of  points  required.  Number  the  points 
1,  2,  3,  4,  from  D  toward  E,  and  from  G  toward  F.  Draw 
lines  from  A  and  B  through  these  points,  and  those  passing 
through  like-numbered  points  meet  in  points  on  the  arc  ACB. 
To  construct  a  point  on  the  curve  below  AB  lay  off  GO  equal 
to  Gl  on  the  arc  FG,  and  D9  equal  to  Dl.  Lines  through  0 
and  B,  and  A  and  9  meet  in  K,  a  point  on  the  curve. 


38 


GEOMETRICAL   PROBLEMS. 


Prob.  7<S.      On   a  chord  AB  to  construct  the  supplementary 
arc  to  ACB,  without  using  the  centre,  C  being  a  point  on  the  arc. 

Join  AC,  AB,  and  BC. 
Make  BAD  and  ABF  each 
equal  to  ACB.  With  A  and 
B  as  centres,  and  any  radius, 
draw  arcs  HD  and  KF.  Di- 
i  vide  the  arcs  DH  and  KF 
>  into  the  same  number  of  equal 
parts  (say  five)  by  the  points 
1,  2,  3,  etc.  Number  the 
points  as  shown.  Draw  lines  from  A  through  1,  2,  3,  etc.  to 
meet  lines  through  the  same  numbers  drawn  from  B.  The  lines 
through  like  numbers  meet  in  points  on  the  required  arc. 

Prob.  79.      To  construct  any  number  of  tangential  arcs  of 
circles,  having  a  given  diameter. 

Suppose  three  (four)  arcs  are  re- 
quired. Upon  the  given  diameter 
as  a  base  draw  an  equilateral  triangle 
ABC  (square  ABCD).  With  each 
vertex  as  a  centre,  and  a  radius  of 
half  a  side,  draw  arcs  of  circles  tangent  to  each  other,  as  shown. 


Prob.'  80.     At  a  point  C  on  a  line  AB  to  draw  two  arcs  of 

circles  tangent  to  AB,  and  to  tiro  parallels   AD  and  BE.  form- 
ing an   arch. 

Make  AD  equal  to  AC,  and  BE  equal  to  BC. 
Ai  C  make  CG  perpendicular  to  AB,  and  at  Dand 
E  draw  the  perpendiculars  DF  and  GE,  meeting 
CG  in  F  and  G,  the  required  centres. 


GEOMETRICAL   PROBLEMS. 


39 


Prob.  81.  Jb  rfraw  an#  number  of  equidistant  parallel 
straight  lines  between  two  given  parallels,  AB  and  CD. 

, a fj  Draw  any  line  A5  at  any  angle  to  AB, 

rj^f and  lay    off  on   it  any  convenient  dis- 

2>--'  y __"    tance  Al   as  a  unit  as  many  times  as 

-  lines  required,  plus  one.  Thus,  if  four 
parallels  are  required,  lay  off  Al  five 
times.  AVith  A  as  a  centre  and  A5  as 
radius  draw  an  arc  cutting  CD  in  point  e.  Join  Ae.  AVith 
centre  A  and  radii  Al,  A2,  A.'>,  etc.,  cut  Ae  in  a,  b,  c,  etc. 
Through  a,  b,  c,  otc.  draw  parallels  to  AB  and  CD. 

Prob.  82.      To  draw  through  a  point  C  a  line  to  meet  the 

inaccessible  intersection  of  two  lines,  AB  and  DE. 

A             b  From  C  draw  any  lines  CA 

p f 

yf  ~T  and  CD   to    the    given  lines. 

-r — |-f  Join  AD,  AC,  and  CD.  Make 
V*  '  BE  parallel  to  AD,  BF  par- 
allel to  AC,  and  EF  parallel 
to  CD.  The  intersection  of  EF  and  BF  gives  a  point  F  in 
the  required  line.     Draw  through  C  and  F. 

Prob.  83.  To  draw  a  perpendicular  to  a  line  AB,  which 
shall  pass  through  the  inaccessible  intersection  of  two  lines,  AE 
and  CD. 

Produce  AB  to  cut  the  lines  in  A  and  C. 
From  A  draw  a  perpendicular  to  CD,  and 
from  C  a  perpendicular  to  AE  ;  these  per- 
pendiculars meet  in  O.  Draw  BO  perpen- 
dicular to  AB.  OB  passes  through  the 
intersection  of  CD  and  AE. 


-^- 


40 


( l  E<  >M  KTIUCAL    PK<  >BLEMS. 


PROB.  84.      To  draw  an  involute  of  a  square  ABCD. 

Produce  the  sides  as  shown.  With 
centre  C  and  radius  CD  draw  the  arc  DE. 
With  centre  B  and  radius  BE  draw  the  arc 
EF.  With  centre  A  and  radius  AF  draw 
the  arc  FG,  etc. 

Note.  Suppose  a  line  to  be  wrapped 
around  and  in  the  direction  of  the  peri- 
meter of  any  plane  figure.  Let  the  line  be  unwound,  keeping 
it  always  straight  in  the  process  of  unwinding.  Any  point  in 
the  line  describes  an  involute.  The  involute  of  polygons  is 
composed  of  arcs  of  circles,  as  in  Prob.  84. 


Prop..  85      To  draw  the  involute  of  a  circle. 

Divide  the  circumference  into  any  nura- 
/  ber  of  equal  parts,  as  at  A,  B,  C,  D,  etc, 
and  draw  radii  to  these  points.  At  A,  B, 
C,  D,  etc.  draw  tangents.  Let  the  curve 
start  at  A.  On  the  tangent  at  B  lay  off  a 
distance  from  B  to  1  equal  to  one  of  the 
parts  into  which  the  circumference  is  di- 
vided. On  the  tangent  at  C  lay  off  a  distance  equal  to  two 
parts  to  2.  On  the  tangent  at  D  three  parts  to  3  ;  from  E  four 
parts  to  4,  etc.  The  curve  through  these  points,  1,  2,  3,  4,  etc., 
is  the  involute  of  the  circle. 


GEOMETRICAL    PROBLEMS. 


41 


Prob.  8G.      To  draw  a  spiral  com  posed  of  semicircles,  the 
radii  being  in  arithmetical  progression. 

Draw  an  indefinite  line,  BAG.  On  the  line 
take  any  two  points,  A  and  B,  as  centres. 
With  A  as  a  centre  and  radius  AB  draw  a 
semicircle.  With  B  as  a  centre  and  radius  BC 
draw  a  semicircle,  and  so  on,  using  A  and  B  as 
centres,  and  taking  the  radii  to  the  end  of  the 
diameter  of  the  last-drawn  semicircle. 


Prob.  87.      To  draw  a  spiral  composed  of  semicircles,  whose 
radii  shall  be  in  geometrical  progression. 

Let  the  ratio  be  2.  Let  AB  he  the  radius  of 
the  first  circle,  A  its  centre.  Draw  the  semicir- 
cle BLC.  With  B  as  a  centre  and  radius  BC 
draw  the  semicircle  CMD.  With  C  as  a  cen- 
tre and  radius  CD  draw  the  semicircle  DNE. 
D  is  the  next  centre,  the  diameter  DE  of  the  last-drawn  circle 
becoming  the  radius  for  the  next  circle.     So  proceed. 


Prob.  88.      To  draw  a  spiral  of  one  turn  in  a  given  circle. 

Divide  the  circle  into  any  num- 
ber of  equal  parts,  say  twelve,  by 
the  lines  OA,  OB,  OC,  etc.,  and  a 
radius  OA  into  the  same  number 
of  equal  parts  by  the  points  1,  2, 
'°  .'5,  4,  etc.  With  O  as  a  centre 
and  radius  01  draw  an  arc  cutting 
OBinM;  with  centre.  ()  and  radius 

02  an  arc  cutting  OC  in  N,  radius 

03  an  arc  cutting  01)  in  point  P, 
etc.     Through  INI,  N,  P,  etc.  draw  the  curve. 

Note.     This  is  "the  spiral  of  Archimedes." 


42 


GEOMETRICAL    PROBLEMS. 


Prop..  .SO.      T<>  draw  in  a  given  circle  a  spiral  of  any  num- 
ber of  turns,  sag  tiro. 

Draw  radii  dividing  the  circle 
into  any  number  of  equal  parts. 
Divide  any  radius,  as  OA,.into  as 
many  equal  parrs  as  turns  in  the 
spiral,  and  divide  each  part  into 
as  many  equal  portions  in  points 
1,  2,  .'!,  4,  etc.  as  the  circle  is  di- 
vided into.  With  centre  0  and 
01,  02,  03,  etc.  as  radii,  draw- 
arcs  to  meet  the  radii  OB,  OC, 
OD,  etc.  in  points  of  the  required  curve. 


PrOB.  90.      Given  the  axes  AB  and  CD  of  an  ellipse  to  draw 
the  curve,  and  at  any  point  on,  the  curve   to  draw  a  tangent. 

Place  the  axes  AB  and  CD 

at  right  angles  to  and  bisecting 

<  each  other  at  0.     "With  centre 

C  and  radius  OA  cut  AB  in 


F  and  F',  which  are  the  foci. 
Between  0  and  F',  or  F,  take 
any  point  G,  dividing  AB  into 
two  parts.  With  centres  F  and 
F  and  radius  AG  draw  arcs  on  either  side  of  AB.  With  the 
same  centres  and  radius  BG  draw  arcs  intersecting  those  drawn 
with  radius  AG,  at  points  L,  M,  N,  and  P,  which  are  points  on 
the  curve.  Take  any  other  point,  T  on  AB,  and  repeat  the 
above  operation  ;  and  so  on  until  as  many  points  as  are  neces- 
sary are  found.  Through  the  points  draw  the  curve.  FM 
plus  F'M  equal  AB.  Let  M  be  the  point  at  which  the  tangent 
is  required.  Produce  FM  and  draw  the  bisector  of  the  angle 
SMF'.     MR  is  the  tangent  required. 


GEOMETRICAL   PROBLEMS. 


43 


Note.  For  drawing  the  ellipse  and  similar  curves  through 
a  series  of  points  the  so-called  French  Curves  are  to  he  used. 

Note.  The  major  axis  is  sometimes  called  the  transverse, 
and  the  minor  the  conjugate,  axis. 

Prob.  91.  To  draw  an  ellipse  by  means  of  a  trammel,  the 
axes  being  given. 

Let  the  semi-axes  be  OA  and  OB.  Mark 
off  on  the  straight  edge  of  a  slip  of  paper 
or  card  MP  equal  to  OA,  and  NP  equal  to 
OB.  Keep  the  trammel  with  the  point  N 
always  on  the  major  axis,  aud  the  point  M 
on  the  minor  axis,  and  P  will  be  a  point  in 
the  curve.  Fmd  as  many  points  as  necessary,  and  draw  the  curve. 


— S 


Prob.  92.      To  draw  an  ellipse,  having  given   the  axes. 


M 


B 

0 

Let  the  semi-axes  be  OA  and  OB.  With 
radii  OA  and  OB  and  centre  O  draw  circles. 
Draw  any  radii,  OM,  ON,  etc.  Make  MP, 
NT,  etc.  perpendicular  to  OA,  and  HP.  KT, 
etc.  parallel  to  OA.  P,  T,  etc.  are  points 
on  the  curve. 


93.      To  draw  an  ellipse,  having  given    the  axes. 

Place  the  axes  at  right  angles  at  their 
centres,  and  on  them  construe!  a  rectangle, 
one  half  being  shown  in  BDEC.  Divide 
OA  and  DA  into  the  same  number  of 
equal  parts  by  points  1,  2,  3,  etc.  Draw 
lines  through  C  and  1,  2,  .">,  etc.  to  meet 
lines  from  B  drawn  to  1,  2,  3,  etc.  on  A  I). 


£ 


P,  Q,  11,  etc.  are  points  on  the  curve. 


44 


G EOMETRI C A L    PROBLE M S . 


Prob.  94. 


To  draw  a  curve  approximating  to  an  ellij>s<>. 

Draw  the  squares  ABDC  and  BP^FD,  and 
their  diagonals,  intersecting  in  G  and  II. 
With  centres  G  and  H  and  radius  GA  draw 
the  arcs  AC  and  EF.  With  centres  B  and 
1)  and  radius  DA  draw  the  arcs  CF  and  AE. 


Prob.  95.      To  draw  on  a  given  line,  AB  as  a  major  axis,  a 
curve  approximating  an  ellipse. 

Divide  AB  into  three  equal  parts  (Prob. 
11)  by  points  C  and  D.  With  centres  C  and 
D  and  radius  CA  draw  two  circles  intersect- 
ing in  E  and  F.  Through  C  and  D  draw 
EGG,  EDII,  FCK,  and  FDL,  meeting  the 
circles  in  points  G,  H,  K,  and  L.  With  centres  E  and  F  and 
radius  EG  draw  the  arcs  GH  and  KL,  completing  the  curve. 


Prob.  96.      Given  the  major  and  minor  axes  of  an  ellipse,  to 
draw  the  curve  approximately. 

Let  C A  be  the  semi-major  axis,  and  BC  the 
semi-minor  axis.  Join  A  and  B.  Make  CD 
equal  to  BC,  and  BE  equal  to  AD.  Bisect 
AE  by  a  perpendicular,  meeting  BC  pro- 
duced in  F.  With  centre  F  and  radius  FB 
draw  the  arc  BH,  and  with  centre  G  and 
radius  GH  draw  the  arc  HA. 
Note.  One  quarter  of  the  whole  curve  is  only  shown,  leav- 
ing to  the  student  the  construction  of  the  full  ellipse. 


GEOMETRICAL   PROBLEMS.  45 

Prob.  97.     Having  the  axes  given,  to  draw  a  curve  of  tan- 
gential arcs  of  circles  approximating  to  the  ellipse. 

-J^s^  AO  is  the  semi-major  axis,  and  OB 

the  semi-minor  axis.     Draw  a  rectangle 


T         <  \      with  the  axes  as  sides.     AOBC  is  one- 

*         •  \ 

r-'    "i     >  n    I  \    quarter  of  the  rectangle.     Draw  AB. 

\"V*f  ~£  From  C  draw  CMP  perpendicular  to 

\  *  [  AB,  and  meeting  BO  produced  in   P. 

\'»j  Make  OE  equal  to  OB.     On  AE  as  a 

ip  diameter  draw  a  semicircle  AKE.    Pro- 

duce OB  to  K.  Make  OL  equal  to  BK.  With  centre  P  and 
radius  PL  draw  the  arc  LN.  Make  AD  equal  to  OK,  and 
with  centre  M  and  radius  MI)  draw  the  arc  DN,  meeting  LN 
in  N.  Draw  NMR  and  PNS.  With  centre  M  and  radius  MA 
draw  AR;  with  centre  N  and  radius  NR  draw  RS,  and  with 
centre  P  and  radius  PS  draw  an  arc  from  S  through  B.  Repeat 
in  each  of  the  quadrants. 


Prob.  98.      To  draw  a  parabola  when  the  abscissa,  AB  and 

the  ordinate  BC  are  given. 

p / t £.__c        Draw  the  rectangle  ABCD,  and  divide 

I*--*'- -j^JU^r-^    ;   ad  and   DC   into^the   same  number  of 

r^v-^'''''  '    equal  parts.    Through  the  points  of  divis- 

%■?'' i    ion   on   AD  draw  parallels   to  AB,  and 

a  Bi 

',\v  I   from  A  draw  lines  to    the  points  on  DC. 

^^~\^^  ;    The   first  line  above  AB  meets  the  line 

■ _Trr^r-r^j    from  A  to  the  first  point  of  division  from 

I)  in  a  point  P  on  the  curve.     The  second  parallel  to  A  meets 

the  second  from  A  to  DC  and  so  on.      P,  Q,  R,  and  C  are  points 

in  the  curve.      Repeat  the  same  below  AB. 


46 


(;]<:<  (METRICAL   PROBLEMS. 


Prob.  99.  To  draw  a  parabola  when  the  directrix  AC  and 
the  focus  D  are  given,  and  to  draw  a  tangent  at  any  point  L  on 
the  curve. 

This  curve  is  such  that  its  apex  E 
is  always  half  way  between  A  and  D, 
and  the  distance  from  D  to  any  point 
upon  the  curve,  as  F,  is  always  equal 
to  the  horizontal  distance  from  F  to 
the  directrix.  Thus  DF  equals  FG, 
and  DH  equals  HK,  etc.  Through 
D  draw  BDA  perpendicular  to  AC. 
This  is  the  axis  of  the  curve.  Draw  parallels  to  AC  through 
any  points  in  AB,  and  with  centre  D  and  radii  equal  to  the  hori- 
zontal distances  of  these  parallels  from  AC  cut  the  correspond- 
ing verticals,  which  will  give  points  on  the  curve. 

To  draw  the,  tangent  at  L.  Draw  the  ordinate  LN,  meeting 
AB  in  N.  Produce  BA  to  the  right.  Make  ET  equal  to  EN. 
Draw  LT,  the  tangent  required. 


Prob.  100.      To  draw  an  hyperbola  when  the  diameter  AB, 
the  abscissa  BC,  and  the  double  ordinate  DE  are  given. 

Complete  the  rectangle 
BCDF,  and  divide  CD  and 
DF  each  into  the  same  num- 
ber of  equal  parts.  Draw 
BL,  BM,  and  B>\  inter- 
secting lines  AK,  AH.  and 
AG  respectively  in  points 
on  the  curve.  Repeat  below 
and  in  the  other  half  of  the 
curve  as  indicated. 


GEOMETRICAL   PROBLEMS. 


47 


To  draw  an  oval  on  the  diameter  of  a  given 

Let  AB  be  the  diameter.     Draw  the  circle 
sn  ACB.    Make  OC  perpendicular  to  AB.    Draw 
the  lines  BCD  and  ACE  indefinitely.     With 
t   centres  A  and  B  and  AB  as  a  radius  draw  the 
arcs  BE  and  AD.     With  centre  C  and  radius 
CD  draw  the  arc  DE. 

Note.  The  centres  A  and  B  may  he  taken  anywhere  on 
the  line  AOB  produced. 

Prob.  102.       Upon  a  given  line  AB  to  draw  an  oval. 

Bisect  AB  at  C,  and  draw 
the  perpendicular  CD.  With 
B  as  a  centre  and  radius  AB 
describe  an  arc  AD.  Bisect 
the  quadrant  AE  ivi  F. 
Through  F  draw  CFG.  AG 
is  the  first  part  of  the  curve. 
Bisect  CB  in  II,  and  draw 
HD.  K  is  the  second  centre. 
Bisect  EL  in  M,  and  draw  KMN.  GN  drawn  from  K  is  the 
second  part  of  the  curve.  Bisect  CH  in  O,  and  draw  DO.  P  is 
the  third  centre.  From  P  through  E  draw  PET.  NT  is  the 
third  part  of  the  curve.  From  E  with  a  radius  ET  carry  the 
curve  to  the  line  DC,  and  repeat  the  operation  for  the  other 
half  of  the  curve.  Draw  a  semicircle  on  the  diameter  AB  for 
the  other  part  of  the  oval. 

The  cycloid  is  the  path  described  by  any  point  in  the  circum- 
ference of  a  circle  which  rolls  along  a  straight  line. 

An  epicycloid  is  the  patli  described  by  any  point  in  the  cir- 
cumference of  a  circle  which  rolls  along  the  outside  of  another 
circle. 


%^D 


48 


GEOMETRICAL    PROBLEMS. 


A  liypocycloid  is  the  path  described  by  any  point  in  the  cir- 
cumference of  a  circle  which  rolls  along  the  inside  of  another 
circle. 

The  rolling  circle  is  called  the  generatrix,  or  generating  cir- 
cle, and  the  line  (straight  or  curved)  on  which  it  rolls  is  called 
the  directrix. 

Every  point  in  the  tire  of  a  wheel  which  rolls  along  the 
ground  in  a  straight  line  describes  a  cycloid. 

Hence  it  is  easily  seen  that  in  one  revolution  or  turning 
around  of  the  circle,  or  wheel,  the  circumference  will  roll  out 
into  a  straight  line.  To  lay  oif  the  circumference  on  the  straight 
line,  either  calculate  its  length  or  divide  the  circle  into  equal 
ares,  and  lav  off  on  the  straight  line  as  many  divisions  as  there 
are  in  the  circle;  each  division  on  the  straight  line  being  equal 
to   l lie  length  of  one  division  on  the  arc. 

The  arc  is  more  than  the  chord,  so  a  distance  greater  than 
the  chord  must  lie  taken,  and  this  is  obtained  by  judgment  or 
by  approximation.  By  dividing  the  arc  into  a  number  of  smaller 
arcs,  so  that  the  chord  practically  coincides  with  the  arc.  the 
small  chord  being  laid  off  as  many  times  as  there  are  small  arcs, 
a  length  of  straight  line  is  obtained  very  nearly  equal  to  the 
arc  given. 


Pkob.  103.      To  construct  a  cycloid. 

"c  Let  AB  be  the  directrix,  and  AD 
the  generating  circle.  Divide  the 
rolling  circle  into  any  number  of 
equal  parts,  say  12,  and  lay  off 
these  lengths  of  arcs  along  AB, 
~->_i  <\  *%  "  *  "IT  giving  points  a,  b,  c^etc.  Through 
1,  the  centre  of  the  generating  cir- 
cle, draw  a  line  parallel  to  AI>. 
This    is    the   line    of  centres.      On 


( I  E<  (METRICAL   PROBLEMS. 


49 


this  lay  off  12,  2.">,  .'!4,  etc.,  equal  to  Aa,  ab,  etc..  and  with 
centres  2,  3,  etc.  draw  the  generating  circle  in  all  its  posi- 
tions tangent  to  AB  at  points  a,  b,  c,  etc.  Draw  through  the 
points  of  division  on  the  rolling  circle  parallels  to  AB,  to  meet 
the  different  positions  of  the  rolling  circle  in  points  P,  R,  S,  T, 
and  U.  These  parallels  are  drawn  in  the  figure  from  points 
p,  r,  s,  t,  and  u.  Repeat  the  process  for  the  other  half  of  the 
curve. 

Another  method  is  to  take  the  chords  of  the  arcs  By?,  Br,  Us, 
etc.,  and  with  centres  a,  b,  c,  etc.  cut  the  respective  circles  in 
points  P,  R,  S,  etc.  The  chord  aP  equals  By?;  bR  equals  Br; 
cS  equals  Bs,  etc. 

Note.  When  the  number  of  divisions  of  the  rolling  circle 
is  large  the  curve  may  be  drawn  by  arcs  of  circles  by  taking  a 
as  a  centre,  and  radius  a  A,  and  drawing  from  A  to  P.  Pro- 
duce Pa  and  Rb  to  meet,  giving  the  centre  for  arc  PR  ;  lib  and 
Sc  meet  at  the  centre  of  arc  Rs,  etc. 


PrOB.  104.      To  construct  an  exterior  epicycloid. 

Let  R-A  be  the  rolling  circle  on 
the  outer  circumference  of  the  direct- 
ing circle.  Divide  R-A  into  any 
number  of  equal  parts  (say  12),  and 
lay  off  these  parts  on  Aab,  etc., 
giving  points  a,  b,  c,  d,  etc. 

With  the  centre  of  the  directing 
circle  as  a  centre,  draw  an  arc  from 
R  giving  the  line  of  centres  R123 
etc.  Draw  from  the  centre  of  the 
directing  circle  radial  lines  through 
a,,  b,  c,  d,  etc.,  meeting  the  line  of  centres  in  points  I,  2,  .'!,  etc., 
the  centres  of  the  different  positionsof  the  rolling  circle.     With 


50  GEOMETRICAL    PROBLEMS. 

centres  1,  2,  3,  etc.  and  radius  RA  draw  the  several  positions 
of  the  rolling  circle.  With  the  centre  of  the  directing  circle 
as  a  centre,  draw  arcs  through  the  points  of  division  of  the  cir- 
cle R-A  to  meet  the  several  positions  of  the  rolling  circle  in 
points  C,  D,  E,  F,  etc.,  which  are  points  on  the  curve.  Draw 
through  C,  D,  E,  F,  etc. 

The  points  of  the  curve  may  be  obtained  by  drawing  from  a 
as  a  centre  and  radius  equal  to  the  chord  of  one  division  of  R-A 
an  arc  to  meet  the  second  position  of  the  rolling  circle  in  C ; 
from  b  with  radius  equal  to  the  chord  of  two  divisions  an  arc 
to  meet  the  third  position  in  D ;  from  c  with  radius  equal  to 
the  chord  of  three  divisions  to  meet  the  fourth  position  in  E, 
etc.  Or  from  a  lay  off  on  the  rolling  circle  tangent  at  a  one 
part  of  R-A  (in  this  case  T\r)  ;  on  the  one  tangent  at  b  two 
[•arts  of  R-A  ;  on  the  one  tangent  at  c  three  parts,  etc. 


Prob.  105.      To  construct  a  hypocycloid. 

See  the  curve  on  the  interior  of  the  directing  cirele  in  figure 
with  Prob.  104.  Aa,  ab,  be,  etc.  are  equal  parts  of  the  circum- 
ference of  the  rolling  circle.  R123  etc.  is  the  line  of  centres. 
The  process  and  directions  are  the  same  as  for  the  epicycloid. 

Note.  When  the  diameter  of  the  rolling  circle  is  equal  to 
the  radius  of  the  directing  circle  the  hypocycloid  becomes  a 
straight  line. 


GEOMETRICAL    PROBLEMS. 


51 


Prob.  106. 


O  lay  off  three  spaces, 
points  of  the  curve. 


To  construct  an  interior  epicycloid. 

Let  the  circle  A-C  roll  on  B-CMN. 
On  the  circumference  of  B-C  lay  off 
equal  arcs  CM,  MN,  NO,  etc.  Draw 
from  M,  N,  O,  etc.  radial  lines 
through  B,  and  make  MD,  NE,  OF 
each  equal  to  the  radius  CA.  With 
centres  D,  E,  F,  etc.  draw  the  cir- 
cles tangent  at  M,  N,  0,  etc.  Lay 
off  Ml  equal  to  CM,  giving  pofnt  1  ; 
from  N  lay  off  two  divisions  each 
equal  to  MC,  giving  point  2 ;  from 
giving  point  3,  etc.      1,  2,  3,  etc.  are 


Prob.  107.      To  draw  a  scroll  for  a  stair-railing. 

The  circle  ABCD  is  the  eye  of  the  scroll. 
Draw  the  diameters  AC  and  BD  at  right 
angles.  Draw  the  chord  AD,  and  bisect 
it  in  6.  Draw  a  line  65  parallel  to  AC. 
Bisect  AE  in  F.  Bisect  EF  in  3.  Make 
E4  equal  to  E3.  Draw  32  and  45  paral- 
lel to  BD,  and  21  parallel  to  AC.  From 
6  draw  an  indefinite  line  parallel  to  BD 
and  produce  65,  21,  etc.  From  point  1 
with  radius  IB  draw  the  arc  BH.  From 
2  and  radius  2H  draw  HJ,  and  so  on.  The  arc  BR  of  the  inner 
curve  is  drawn  from  P  with  radius  PB;  the  arc  RS  is  drawn 
from  6  with  radius  6R. 


52 


CK.n.MKTIMCAL     PROBLEMS. 


Prob.  108. 

A 


Another  method. 


Suppose  the  scroll  to  come  between 
the  outside  of  the  rail  AB  and  the  line 
CD.  Draw  A9  perpendicular  to  AB 
and  CD,  and  divide  it  into  nine  equal 
parts.  On  the  sixth  division  as  a  side 
draw  the  square  etbcQ.  With  centre  a 
and  radius  aA  draw  the  quadrant  AE. 
With  centre  6  and  radius  6E  draw  arc 
EF.  With  centre  c  and  radius  cFdraw 
FG  ;  with  centre  6  and  radius  0>G  draw  arc  GH ;  with  centre  e 
and  radius  eH  draw  arc  HK  ;  and  with  centre/ and  radius /K 
draw  the  arc  KL.  For  the  inner  curve  start  with  a  as  a  centre 
and  radius  a\  and  proceed  as  lor  the  outer. 


Prob.  109. 
being  given. 


To  construct  a  spired,  its  greatest  diameter  AB 


Divide  AB  into  eight  equal  parts 
by  points  1,  2,  3,  etc.  On  45  as  a 
diameter  draw  a  circle  CDEF.  This 
circle  is  the  eye  of  the  spiral.  In- 
scribe a  square  CDEF,  as  shown  in 
the  enlarged  drawing  of  the  eye.  Draw 
the  central  diameters  of  the  square. 
1-2.10  and  11,9.  Divide  these  diam- 
eters into  six  equal  parts,  and  number 
as  shown.  With  1  as  a  centre  and  ra- 
dius 1H  draw  an  arc  HK  to  meet  a  horizontal  produced  from 
2  through  1.  With  2  as  a  centre  and  radius  2K  draw  the  arc  KL 
to  meet  a  vertical  through  3  and  2.  With  centre  3  and  radius  3L 
draw  arc  LM  meeting  a  horizontal  produced  through  -1  and  3. 
With  4  as  a  centre  and  radius  4M  draw  arc  MN  to  meet  a  line 
drawn  through  5  and  4  at.  N,  and  so  proceed. 


GEOMETRICAL   PROBLEMS. 


53 


The  curve  may  be  commenced  by  taking  12  as  a  centre  and 
radius  12  A  and  drawing  from  without  toward  the  centre. 


Prob.  110.      To  describe  an  Ionic  volute. 

Let  AB  be  the  vertical 
measure  of  the  volute.  Di- 
vide AB  into  seven  equal 
parts,  and  from  C,  the  lower 
extremity  of  the  fourth  di- 
vision, draw  CF  perpendic- 
ular to  AB,  of  indefinite 
length.  From  any  point  on 
CF  as  a  centre,  with  a  ra- 
dius equal  to  one-half  of 
one  of  the  divisions  of  AB, 
draw  the  circle  HIJK,  form- 
ing the  eye  of  the  volute.  Draw  the  diameter  HJ  perpendicu- 
lar to  CF.  Draw  the  square  HIJK,  bisect  its  sides,  and  draw 
the  square  12  L  M  11.  Draw  the  dividing  lines  of  the  square 
as  shown  in  the  smaller  figure,  and  extend  them.  The  divis- 
ions corresponding  to  12  N  are  equal.  The  divisions  OP  and 
RL  are  each  equal  to  one-half  of  12  N.  From  1  as  a  centre 
and  radius  1H  draw  the  arc  US  ;  from  2  as  a  centre  and  radius 
2S  the  arc  ST ;  from  centre  3  and  radius  8T  the  arc  TU ;  and 
so  proceed  in  the  order  of  the  numerals. 

In  drawing  the  inner  curves  the  dots  on  the  diagonals  in  the 
small  figure  indicate  the  centres.  The  division  of  the  square, 
of  which  12  N  is  one  side,  shows  how  these  centres  are  found. 


54 


GEOMETRICAL   PROBLEMS. 


Fillet 


(  Bead 


Torus 


Cyma  Recta 


Cyma  Reversa 


The  Roman  moldings  are  given  above,  the  method  of  con- 
struction being  evident.  All  the  arcs  are  arcs  of  circles,  and 
the  angles  are  45°,  except  in  the  flatter  form  of  the  cyma  recta, 
where  the  line  of  centres  is  at  30°. 


CHAPTER  III. 


14.  It  is  supposed  that  the  student  has  now  become  familiar 
with  the  use  of  the  instruments  necessary  for  the  construction 
of  a  drawing  in  pencil,  and  has  acquired  a  certain  degree  of 
proficiency  in  handling  them  which  is  necessary  for  accurate 
work.  The  next  step  is  to  learn  to  ink  a  drawing  after  it  has 
been  pencilled. 

Starting  with  a  good  pen,  in  good  condition,  and  a  smooth, 
well-ground  black  ink,  it  only  remains  for  the  student  to  learn 
to  make  a  clean,  sharp,  even  line.  This  may  seem  at  first  like 
an  easy  thing  to  do  ;  nevertheless,  the  ability  to  make  a  good 
ink  line  every  time  comes  to  most  students  very  slowly,  and 
after  a  great  deal  of  practice.  Therefore,  before  beginning  to 
work  on  the  plates,  which  are  to  be  finished  carefully  and 
handed  in  for  inspection,  it  will  save  a  considerable  time  and 
paper  to  make  lines  against  the  triangle  without  regard  to  their 
length,  direction,  or  location,  until  the  student  is  thoroughly 
familiar  with  his  pen  and  can  make  a  fair  line.  Several  hours, 
if  necessary,  in  this  preliminary  practice  will  be  invaluable. 

1.5.  India  Ink.  A  special  ink,  called  India  ink,  is  always 
used  in  making  drawings.  It  comes  either  in  the  stick  form, 
and   has   to  be  ground  as  used,  or  in  the  liquid  form.     In  the 


56  INKING. 

latter  form  it  is  held  in  solution  by  an  acid  which  corrodes  the 
pen  and  eats  into  the  fibre  of  the  paper,  so  that  if  it  is  desired 
to  erase  a  line  it  is  much  more  difficult  than  if  made  with  the 
ground  ink.  This  ink  is  also  very  liable  to  rub  off  like  soot. 
The  only  advantage  it  has  is  the  saving  of  time  in  preparation. 
This  kind  of  ink  cannot  be  recommended  for  anything  except 
coarse,  rough  work.     It  should  not  be  used  for  tinting. 

To  prepare  the  stick  ink  for  use  place  a  small  amount  of 
water  in  the  ink  slab  or  saucer  (the  slab  should  be  perfectly 
clean),  then  grasp  the  ink  firmly  and,  with  a  rotary  motion, 
grind  until  the  liquid  is  black  and  a  little  sluggish  in  its  motion. 
After  it  has  reached  the  point  when  it  is  black  enough  the 
grinding  should  cease,  as  a  continuation  only  makes  the  liquid 
thicker,  thus  causing  it  to  flow  less  freely  from  the  pen.  The 
liquid  will  look  black  in  the  slab  after  a  very  little  grinding,  but 
the  necessary  consistency  will  not  be  reached  for  some  time. 
In  ordex  to  determine  when  the  proper  point  is  reached  make 
a  heavy  line  with  the  drawing  pen  on  a  piece  of  paper  and  wait 
for  it  to  dry  ;  do  not  go  over  this  line  a  second  time.  If  the 
ink  has  not  been  ground  sufficiently,  it  will  look  pale  after  it  is 
dry,  in  which  case  more  grinding  is  necessary.  If  at  any  time 
the  ink  becomes  too  thick,  it  can  be  diluted  by  putting  in  more 
water  and  mixing  thoroughly.  The  stick  should  always  be 
wiped  dry  after  using,  to  prevent  its  crumbling.  The  ground 
ink  should  be  kept  covered  as  much  as  possible  to  prevent  evap- 
oration, which  would  soon  cause  it  to  become  too  thick  for  use. 
It  is  not  advisable  to  prepare  a  large  quantity  of  ink  at  once, 
as  the  greater  the  amount  the  longer  it  takes,  and  freshly 
ground  ink  is  preferable.  If  carefully  covered,  however,  it  may 
be  kept  two  or  three  days.  In  case  the  ink  becomes  dry  in  the 
saucer  it  should  all  be  washed  out,  as  it  is  almost  impossible  to 
redissolve  it  entirely  so  that  there  will  not  be  little  scales  which 
get  into  the  pen  and  cause  the  ink  to  flow  irregularly. 


INKING.  57 

16.  Drawing  Pi-:n.  This  instrument,  commonly  called  a 
right-line  pen,  is  one  of  the  most  important  of  the  drawing 
instruments,  and  it  is  very  essential  that  it  be  of  good  quality. 
The  screw  is  used  to  adjust  the  distance  between  the  nibs,  in 
order  to  make  the  line  of  the  desired  weight.  The  ink  may 
be  placed  between  the  nibs  by  means  of  a  brush  or  strip  of 
paper,  but  it  is  more  convenient  to  dip  the  pen  into  the  ink, 
being  careful  to  wipe  the  outside  of  the  nibs  before  using. 

While  inking  the  pen  should  be  held  so  that  both  nibs  rest 
on  the  paper  evenly,  and  it  should  be  inclined  a  little  to  the 
right,  or  in  the  direction  of  the  line,  with  its  flatter  side  against 
the  triangle  or  straight  edge,  the  end  of  the  middle  finger  rest- 
ing on  the  head  of  the  screw.  A  slight  downward  pressure  is 
necessary  (the  greater  the  rougher  the  paper),  but  do  not  press 
against  the  ruler,  as  the  lines  would  be  uneven  in  thickness. 
The  ruler  is  simply  a  guide  for  the  pen.  The  lines  should 
always  be  drawn  from  left  to  right  (relative  to  the  person  and 
not  to  the  drawing).  If  it  is  desirable  to  go  over  a  line  a  sec- 
ond time  for  any  reason,  it  should  be  drawn  in  the  same  direc- 
tion ;  never  go  backward  over  it. 

In  inking  a  curved  line  by  means  of  the  right-line  pen  and 
irregular  curve,  it  is  necessary  to  constantly  change  the  direc- 
tion of  the  pen  so  that  the  nibs  shall  always  be  tangent  to  the 
curve.     This  requires  considerable  practice  to  do  nicely. 

In  case  the  ink  does  not  flow  freely  from  the  pen,  moisten  the 
end  of  the  finger  and  touch  it  to  the  end  of  the  pen,  and  try  it 
on  a  piece  of  waste  paper.  If  this  fails  the  pen  should  be 
wiped  out  cleau  and  fresh  ink  put  in.  In  making  fine  lines  the 
nibs  of  the  pen  are  near  to  each  other,  consequently  the  ink 
dries  between  them  quite  rapidly,  hence  it  will  be  found  advis- 
able to  clean  out  the  pen  thoroughly  quite  frequently  to  insure 
perfect  lines.     This  is  one  of  the  secrets  of  being  ahle  to  make 


58  INKINC. 

good  fine  lines  ;  they  should  also  be  made  more  rapidly  than 
heavy  lines;  the  heavier  the  line  the  slower  the  pen  should  be 
moved.  Do  not  keep  the  point  of  the  pen  too  near  the  straight 
edge,  as  the  ink  is  liable  to  flow  against  it,  thus  causing  a  blot. 
Especial  attention  should  be  given  to  the  care  of  the  pens  ; 
they  should  always  be  carefully  wiped  after  using,  and  should 
not  be  put  away  with  any  ink  dried  on  them,  nor  allowed  to 
get  rusty  on  the  inside  of  the  nibs.  Any  old  piece  of  cotton 
cloth  will  answer  to  wipe  the  pens  and  stick  of  ink  on. 

17.  How  to  Sharpen  the  Pen.  To  make  good  lines 
the  pen  must  be  kept  in  first-class  condition, —  that  is,  not  only 
clean,  but  sharp;  and  every  draftsman  should  be  able  to 
sharpen    bis   own    pen. 

The  curve  at  the  point  of  the  nibs  of  the  pen  should  always 
lie  a  semi-ellipse,  with  its  long  diameter  coinciding  with  the  axis 
of  the  pen  ;  it  should  not  be  a  semicircle,  nor  should  it  be 
pointed.  The  student  is  advised  to  look  carefully  at  the  points 
of  his  new  pen,  so  as  to  get  a  correct  idea  of  the  proper  curve 
before  it  becomes  changed  by  wear.  When  this  curve  becomes 
changed  by  wear,  or  if,  from  any  other  cause,  one  nib  is  longer 
than  the  other,  the  nibs  should  be  screwed  together,  then,  hold- 
ing  the  pen  in  a  plane  perpendicular  to  the  oil-stone,  draw  it 
back  and  forth  over  the  stone,  changing  the  slope  of  the  pen 
from  downward  and  to  the  right  to  downward  and  to  the  left, 
or  vice  versa,  for  each  forward  or  backward  movement  of  the 
pen,  so  as  to  grind  the  points  to  the  proper  curve,  making  them 
also  of  exactly  the  same  length. 

This  process,  of  course,  makes  the  points  even  duller  than 
before,  but  it  is  a  necessary  step.  Next  separate  the  points  a 
little  by  means  of  the  screw,  and  then  place  either  blade  upon 
the  stone,  keeping  the  pen  at  an  angle  of  about  15°  with  the 


INKING.  59 

face  of  the  stone,  move  it  backward  and  forward,  at  the  same 
time  giving  it  an  oscillating  motion,  until  the  points  are  sharp. 
This  is  quite  a  delicate  operation,  and  great  care  should  be  exer- 
cised at  rirst.  The  pressure  upon  the  stone  should  not  be  very 
great,  and  it  is  well  to  examine  the  point  verj'  often  so  as  to  be 
sure  and  stop  when  each  nib  has  been  brought  to  a  perfect 
edge,  otherwise  one  nib  is  liable  to  be  longer  than  the  other 
and  the  pen  will  not  work  well,  even  if  each  nib  is  sharp  of 
itself. 

Although  the  points  want  to  be  brought  to  a  perfect  edge, 
they  should  not  be  sharp  like  a  knife,  as  in  that  case  they  would 
cut  the  paper.  It  will  probably  be  necessary  to  try  the  pen 
with  ink  to  be  sure  that  it  is  in  good  condition.  Sometimes  a 
slight  burr  is  formed  on  the  inside  of  the  blades  ;  this  is  re- 
moved by  separating  the  points  still  farther,  so  as  to  insert  the 
knife-edge  of  the  oil-stone  between  them  and  draw  it  carefully 
through.     One  motion  should  be  sufficient  to  remove  the  burr. 

The  pen  should  never  be  sharpened  by  grinding  the  inside 
of  the  blades  other  than  just  indicated. 

18.  Inking  a  Drawing.  In  inking  a  drawing  it  is  prefer- 
able to  ink  all  the  circles  and  arcs  first,  as  it  is  easier  to  make 
the  straight  lines  meet  the  arcs  than  the  reverse.  Of  a  number 
of  concentric  circles  the  smallest  should  be  inked  first.  Here, 
as  in  the  case  of  the  pencil  compasses,  the  pen  point  should  be 
kept  nearly  vertical,  the  top  of  the  compass  being  inclined  a 
little  toward  the  direction  of  revolution,  and  there  should  be  a 
slight  downward  pressure  on  the  pen  point,  but  none  on  the 
needle  point. 

Where  a  large  number  of  lines  meet  at  a  point  care  should 
be  taken  to  avoid  a  blot  at  their  intersection.  These  lines 
should  be  drawn  from  rather  than  toward  the  point,  and  each 
line  should  be  thoroughly  dry  before  another  is  drawn. 


60  INKING. 

In  case  of  two  lines  meeting  at  a  point  neither  line  should 
stop  before  reaching  the  point,  nor  go  beyond  it.  Either  of 
these  defects  gives  a  very  ragged  appearance  to  the  drawing. 

19.  Stretching  Paper.  For  ordinar}'  small  line  drawings 
it  is  usually  sufficient  to  fasten  the  paper  to  the  board  by  means 
of  thumb  tacks,  but  for  large  drawings,  or  those  which  are  to 
be  tinted  at  all,  it  is  necessary  to  stretch  the  paper  by  wetting 
it  and  fasten  it  to  the  board  with  mucilage.  To  do  this  lay 
the  paper  on  the  board,  fold  over  about  one-half  an  inch  along 
each  edge  of  the  sheet ;  do  not  cut  the  corners ;  next  wet  the 
upper  surface  of  the  paper,  except  that  portion  folded  over ;  do 
not  rub  the  surface  of  the  paper,  simply  press  the  sponge  against 
it  on  all  parts ;  apply  the  mucilage  to  one  of  the  edges  and 
fasten  that  edge  down,  beginning  at  the  middle  and  rubbing 
toward  either  end ;  do  the  same  with  the  opposite  edge  next, 
giving  a  slight  pull  to  the  paper  as  it  is  fastened  down  ;  repeat 
this  process  for  the  two  remaining  edges. 

It  is  very  important  that  the  edges  of  the  paper  where  the 
mucilage  is  to  be  applied  should  be  kept  dry,  so  that  the  muci- 
lage will  be  ready  to  act  as  soon  as  it  can  dry,  and  to  facilitate 
this  the  less  mucilage  you  can  use  and  accomplish  the  result 
the  better.  If  a  large  quantity  of  mucilage  is  used  it  will 
moisten  the  edges  of  the  paper  so  much  that  it  will  be  likely 
not  to  stick,  as  the  body  of  the  paper  will  dry  as  soon  as  the 
edge,  and  therefore  pull  them  up.  The  drying  of  the  mucilage 
can  be  hastened  by  rubbing  the  edge  briskly  with  a  piece  of 
thick  paper  under  the  lingers  until  it  becomes  hot.  The  board 
should  never  be  placed  near  the  fire  or  radiator  to  hasten  the 
drying,  as  it  would  dry  the  paper  before  the  mucilage  set,  caus- 
ing the  edges  to  be  pulled  up.  The  board  should  be  left  to 
dry  in  a  horizontal  position,  and  all  the  superfluous  water  should 


TINTING.  61 

be  removed  with  a  sponge,  so  as  to  avoid  water  marks  in  the 
paper,  which  always  show  in  tinted  drawings.  In  some  cases, 
when  the  mucilage  sets  slowly,  it  may  be  necessary  to  moisten 
the  centre  of  the  paper  sometime  after  stretching,  to  prevent 
its  pulling  up  the  edges  by  drying  too  rapidly. 

20.  Correcting  and  Cleaning  Drawings.  Pencil  lines 
are  removed  by  meaus  of  a  piece  of  rubber.  When  a  mistake 
is  made  in  inking,  or  it  is  desired  to  change  a  completed  draw- 
ing, it  becomes  necessary  to  erase  an  ink  line.  This  can  be  done 
by  means  of  a  rubber  ink  eraser,  the  same  as  in  the  case  of 
pencil  lines,  except  that  much  more  rubbing  is  necessary.  Ink 
lines  can  also  be  removed,  and  more  quickly,  by  means  of  a 
knife ;  in  this  case  care  should  be  taken  not  to  use  the  point  of 
the  knife,  as  V-shaped  holes  are  made  which  will  always  show. 
The  flat  portion  of  the  knife  should  be  used.  After  erasing  an 
ink  liue,  the  surface  which  has  been  made  rough  by  scratching 
should  be  rubbed  down  with  some  hard,  perfectly  clean,  rounded 
instrument  before  inking  other  lines  over  it. 

A  drawing  can  be  cleaned  by  means  of  India  rubber,  or  stale 
bread  crumbled  on  the  drawing  and  rubbed  over  it.  Although 
dirt  can  be  removed  from  a  drawing,  it  should  be  the  aim  of 
the  draftsman  to  keep  it  as  clean  as  possible.  Therefore,  the 
drawing  should  be  kept  covered  when  not  being  worked  upon, 
and,  if  the  drawing  is  a  large  one,  all  except  that  portion  which 
is  in  use  should  be  kept  covered. 

tinting. 

21.  Tinting  may  be  done  in  colors  or  India  ink,  as  desired. 
The  method  of  putting  on  the  tint  is  the  same  in  either  case ; 
consequently,  we  will  take  up  only  the  India-ink  tint  here. 

If  a  drawing  is  to  be  tinted,  the  paper  must  be  stretched  as 


62  TINTING. 

explained  in  the  first  part  of  this  chapter.  Especial  care 
should  be  taken  to  keep  the  paper  perfectly  clean.  That  por- 
tion of  the  drawing  which  is  to  be  tinted  must  not  be  touched 
with  the  India  rubber,  as  the  surface  is  thereby  made  rough 
and  will  not  take  a  uniform  tint.  Hence,  in  laying  out  the 
work  pencil  lines  must  not  be  made  on  the  surface  to  be  tinted. 

22.  Preparation  of  the  Tint.  Clean  the  ink  slab,  water 
glass,  and  the  brushes  thoroughly,  also  be  sure  that  there  are 
no  scales  on  the  stick  of  ink  which  could  possibly  come  off. 
Fill  the  slab  about  half  full  of  water,  and  grind  the  ink  as  pre- 
viously explained  until  it  is  black,  but  not  thick.  Fill  the  water 
glass  about  half  full  of  clean  water,  and  with  a  brush  transfer 
enough  of  the  ink  in  the  slab  to  the  glass  to  make  a  light  tint. 
It  is  hard  to  get  an  ink  which  is  absolutely  free  from  specks, 
therefore,  it  is  well  to  let  the  ink,  after  it  is  prepared  in  the 
slab,  stand  a  short  time  to  allow  these  specks  to  settle  to  the 
bottom  ;  then,  in  transferring  the  ink  to  the  glass,  do  not  plunge 
the  brush  down  to  the  bottom  of  the  slab,  thus  taking  up  this 
sediment,  but  let  the  brush  fill  from  the  surface  of  the  liquid. 

The  mixture  in  the  water  glass  is  the  one  to  be  used  for 
tinting,  and  it  is  better  not  to  make  it  as  dark  as  you  wish  it 
upon  the  drawing  when  finished,  as  it  is  much  easier  to  put  on 
a  light  tint  evenly  than  a  dark  one.  The  required  depth  of 
shade  can  be  obtained  by  successive  washes.  Let  each  wash  dry 
thoroughly  before  putting  on  another.  A  smoother  effect  can 
usually  be  obtained,  especially  on  a  large  surface,  by  going  over 
the  surface  to  be  tinted  with  clean  water  first,  and  then  letting 
it  dry. 

23.  Laying  on  the  Tint.  Having  laid  out  the  surfaces 
to  be  tinted,  incline  the  board  so  as  to  slope  like  an  ordinary 


TINTING.  63 

desk  ;  then  dip  your  brush  into  the  tint  you  have  mixed,  and 
take  up  as  much  of  the  liquid  as  it  will  carry,  begin  in  the  upper 
left-hand  corner  of  the  surface  and  draw  it  along  the  upper . 
boundary,  holding  the  brush  nearly  vertical  and  leaving  quite  a 
puddle  as  you  proceed.  Lead  this  puddle  gradually  downward 
by  going  across  the  surface  from  left  to  right,  about  a  quarter 
of  an  inch  at  a  time,  dipping  the  brush  frequently  in  the  tint 
so  as  to  keep  the  puddle  about  the  same  size  all  the  time  ;  it 
should  not  be,  large  enough  to  run  down  at  any  point.  This 
puddle  should  not  be  left  standing  at  any  place  any  longer  than 
is  absolutely  necessary,  as  it  is  very  apt  to  leave  a  streak  ; 
therefore,  having  commenced  on  a  surface,  finish  it  as  quickly  as 
possible, —  do  not  let  anything  interrupt  you. 

When  you  get  to  the  bottom  dry  your  brush  on  a  piece  of 
blotting  paper,  and  with  the  dry  brush  take  up  the  superfluous 
tint,  as  you  would  with  a  sponge,  until  it  looks  even. 

Iu  laying  on  the  tint  do  not  bear  on  with  the  brush,  as  the 
brush  marks  would  be  liable  to  show,  but  use  the  point  only, 
just  touching  it  to  the  paper  so  as  to  wet  it,  and  the  tint  will 
follow  along  of  itself  (the  board  being  properly  inclined). 

In  following  the  boundaries  of  the  surface  to  be  tinted  let 
the  brush  be  pointed  towards  the  boundary  from  the  inside  of 
the  surface. 

This  gives  what  is  called  a,  flat  tint,  and  is  used  to  represent 
surfaces  which  are  parallel  to  the  plane  of  projection. 

24.  For  representing  surfaces  which  are  oblique  to  the  plane 
of  projection  a  graduated  tint  is  necessary.  There  are  two 
methods  of  doing  this, —  the  French  and  the  American. 

The  French  method  consists  in  dividing  the  surface  into  small 
divisions  (these  divisions  should  be  indicated  only,  not  drawn 
across  the  surface),  laying  a  flat  tint  on  the  first  space,  and 
when  this  is  dry  laying  another  flat  tint  on  the  first  two  spaces. 


64  TINTING. 

Proceed  iu  this  way  until  the  whole  surface  is  covererl,  com- 
mencing at  the  first  space  each  time.  By  this  method  the 
shading  shows  streaks  of  tint  of  different  depths,  but  are  almost 
unnoticeable  if  the  divisions  are  taken  quite  small. 

The  American  method  is  most  used,  aud  is  called  shading  by 
softened  tints. 

There  are  two  ways  of  doing  this  :  — 

1st.  By  mixing  a  small  amount  of  dark  wash  at  first,  and 
starting  as  if  you  were  to  put  on  a  flat  tirft,  and  then,  by  re- 
peated additions  of  clean  water,  going  over  a  little  more  surface 
at  each  addition,  gradually  make  the  dark  tint  lighter  until  you 
are  using  almost  pure  water. 

2nd.  Divide  the  surface  into  divisions,  as  in  the  French 
method,  only  not  so  many  ;  put  on  a  flat  medium  tint  on  the 
first  space,  but,  instead  of  taking  up  all  the  tint  from  the  bot- 
tom edge  of  the  surface,  leave  a  slight  amount,  touch  the  brush 
to  some  clean  water  and  apply  it  to  the  lower  edge  of  the  pud- 
dle, thus  making  a  lighter  tint,  and  bring  down  this  new  tint  a 
short  distance  ;  repeat  this  a  few  times  until  the  tint  has  prac- 
tically no  color.  Be  careful  to  remove  the  most  of  the  tint  from 
the  brush  each  time  before  touching  it  to  the  clear  water.  This 
work  must  be  done  even  quicker  than  the  ordinary  flat  tint. 
Use  as  little  tint  or  water  in  the  brush  as  you  can  and  not 
have  it  dry  in  streaks.  Let  this  dry,  and  then  repeat  the  proc- 
ess, commencing  at  the  top  and  going  over  two  spaces  with  the 
flat  tint  and  softening  off  the  lower  edge,  and  so  on,  commenc- 
ing at  the  top  each  time.  Usually  the  tint  should  be  softened 
out  in  the  length  of  one  division.  If  this  shading  is  done  per- 
fectly, there  will  be  a  gradual  change  in  the  tint  from  beginning 
to  end. 


CHAPTER  IV. 


PROJECTIONS. 


25.  Orthographic  Projection,  or  Descriptive  Geometry, 
is  the  art  of  representing  a  definite  body  in  space  upon  two 
planes,  at  right  angles  with  each  other,  by  lines  falling  perpen- 
dicularly to  the  planes  from  all  the  points  of  the  intersection  of 
every  two  contiguous  sides  of  the  body,  and  from  ail  points  of 
its  contour. 

26.  These  planes  are  called  coordinate  planes,  or  the  planes 
of  projection,  one  of  which  is  horizontal,  and  the  other  vertical. 
H  and  V,  Fig.  1,  represent  two  such  planes  and  their  line  of 
intersection  GL  is  called  the  ground  line. 

27.  We  shall  only  take,  in  this  book,  just  enough  of  the 
elementary  principles  of  projections  to  enable  the  student  to 
make  working  drawings  of  simple  objects. 

28.  Since  solids  are  usually  made  up  of  planes,  planes  of 
lines,  and  lines  of  points,  if  we  thoroughly  understand  the  prin- 
ciples involved  in  the  projections  of  points,  we  ought  to  be  able 
to  draw  the  projections  of  lines,  of  planes,  and  of  solids.  The 
only  difference  being  that,  with  a  large  number  of  points,  the 
student  is  liable  to  get  them  confused.  To  avoid  this  liability 
it  is  advisable,  at  first,  to  number  the  points  of  a  solid  and  their 
corresponding  projections  as  fast  as  found  lightly  in  pencil,  and 
erase  them  after  the  problem  is  finished.     Do  not  try  to  draw 


66  PROJECTIONS. 

the  object  all  at  once,  it  is  impossible ;  one  point  at  a  time  is 
all  that  can  be  drawn  by  anybody,  and  in  this  way  the  most 
complicated  objects  become  simple,  even  though  it  may  take  a 
long  time  to  complete  the  drawing. 

29.  The  projection  of  any  point  in  space  on  a  plane  is  the 
point  at  which  a  perpendicular  drawn  from  the  given  point  In 
the  plane  pierces  the  plane. 

This  perpendicular  is  called  the  projecting  line  of  the  point. 
Thus,  in  Fig.  1,  ah  is  the  projection  of  the  point  a  on  the  plane 
H,  and  o*  of  the  same  point  on  the  plane  V.  These  are  called 
respectively  the  horizontal  and  vertical  projections  of  the  point 
a  ;  aah  is  called  the  horizontal  projecting  line  of  the  point  a, 
and  aav  the  vertical  projecting  line  of  the  same  point. 

The  horizontal  projecting  line  aah  is  perpendicular  to  H,  by 
definition,  the  plane  V  is  assumed  perpendicular  to  H,  hence 
aah  is  parallel  to  the  plane  V,  and  aav  is  equal  in  length  to 
ahb.  Also  the  vertical  projecting  line,  for  the  same  reason,  is 
parallel  to  H,  consequently  aah  is  equal  in  length  to  avb. 

From  the  definition  it  is  readily  seen  that  each  point  in  a  line 
perpendicular  to  a  plane  will  have  its  projection  on  that  plane 
in  one  and  the  same  point ;  hence  one  projection  of  a  point 
does  not  definitely  locate  its  position  in  space. 

30.  From  the  preceding  article  the  following  principles  may 
be  noted :  — 

First,  the  perpendicular  distance  from  the  horizontal  projec- 
tion of  a  point  to  the  ground  line  is  equal  to  the  perpendicular 
distance  of  the  point  in  space  from  the  vertical  plane ;  or, 
briefly,  the  horizontal  projection  of  a  point  indicates  the  distance 
of  the  point  in  space  in  front  of  V,  but  it  conveys  no  idea  of  its 
distance  above  H. 

Second,  the  perpendicular  distance  from  the  vertical  projec- 
tion of  a  point  to  the  ground  line  is  equal  to  the  perpendicular 


PROJECTIONS.  67 

distance  of  the  point  in  space  from  the  horizontal  plane ;  or, 
briefly,  the  vertical  projection  of  a  point  indicates  the  height  of 
the  point  in  space  above  H,  but  it  conveys  no  idea  of  its  dis- 
tance in  front  of  V. 

31.  If  from  the  points  a"  and  ah,  Fig.  1,  perpendiculars 
should  be  erected  to  each  coordinate  plane,  they  will  intersect 
at  the  point  a  in  space ;  and  as  two  straight  lines  can  intersect 
at  only  one  point,  there  is  only  one  point  in  space  which  can 
have  ah  and  a®  for  its  projections.  Hence  two  projections  of 
a  point  are  always  necessary  to  definitely  locate  its  position  in 
space. 

32.  It  is  evident  that  it  would  be  very  awkward  to  make 
our  drawings  on  planes  at  right  angles  to  each  other;  hence 
the  vertical  coordinate  plane  is  supposed  to  be  revolved  back- 
ward about  its  line  of  intersection  GL  with  the  horizontal 
plane  until  it  forms  one  and  the  same  surface  with  the  hori- 
zontal plane,  which  may  be  considered  to  be  the  plane  of  the 
paper. 

In  this  revolution  all  points  in  the  vertical  plane  keep  the 
same  distance  from  the  ground  line,  and  their  relative  positions 
remain  unchanged.  Thus,  av  revolves  to  av,  arb  being  equal 
to  a vb,  and  as  a*b  was  perpendicular  to  GL  before  revolution 
it  will  be  so  after  revolution,  and  will  form  one  and  the  same 
straight  line  with  ahb. 

Therefore,  the  two  projections  of  a  point  must  always  be  on 
one  and  the  same  straight  line,  perpendicular  to  the  ground  line. 

Now,  if  we  draw  a  line  across  our  paper  and  call  it  GL,  all 
that  portion  in  front  of  this  line  will  represent  the  horizontal 
plane,  and  that  portion  behind  it  will  represent  the  vertical 
plane,  and  the  point  a  located  as  in  Fig.  1  is  represented  by  its 
projections  on  the  plane  of  the  paper  as  shown  in  Fig.  2. 


68  PROJECTIONS. 

33.  A  point  situated  upon  either  of  the  coordinate  planes  has 
for  its  projection  on  that  plane  the  point  itself  and  its  other  pro- 
jection is  in  the  ground  line. 

This  is  readily  seen  by  referring  to  Fig.  1  ;  cv  and  ch  are  the 
projections  of  a  point  on  H,  and  dv  and  dh  of  a  point  in  V. 
These  points  are  represented  on  the  plane  of  the  paper  as  shown 
by  the  same  letters  in  Fig.  2. 

NOTATION. 

34.  We  will  designate  a  point  in  space  by  a  small  letter,  and 
its  projections  by  the  same  letter  with  an  h  or  v  written  above ; 
thus  ah  represents  the  horizontal  and  av  the  vertical  projection 
of  the  point  a.  This  point  may  be  spoken  of  as  the  point  a, 
or  as  the  point  whose  projections  are  arah. 

35.  The  horizontal  coordinate  plane  will  be  designated  by 
the  capital  letter  H,  the  vertical  by  the  capital  letter  V. 

36.  Construction  lines  are  those  which  are  made  use  of  sim- 
ply to  obtain  required  results.  They  are  not  a  necessary  part 
of  the  drawing,  and  when  left  on  a  drawing  are  intended  to 
show  the  individual  steps  taken. 

To  this  end  the  student  is  expected  to  ink  in  all  construction 
lines  illustrating  the  special  subject  in  hand  until  especially 
directed  not  to  do  so.  That  is,  while  on  the  subject  of  projec- 
tions it  is  not  desired  to  have  the  geometrical  construction  lines 
inked  in,  but  only  those  which  refer  to  projections.  When  on 
the  subject  of  shadows  only  those  which  show  how  the  shadow 
is  found  are  required. 

These  lines  should  be  inked  in  with  a  light,  short  dash  not 
more  than  TJ5  of  an  inch  long,  and  as  light  as  the  student 
finds  he  can  make  easily. 

All  lines  representing  the  projections  of  single  lines,  or  edges 
of  planes  or  solids,  if  visible,  are  inked  in  with  a  full,  continu- 


PROJECTIONS.  69 

cms  line,  a  little  heavier  than  the  construction  lines  ;  if  invisi- 
ble, they  should  be  marie  with  short  dashes  the  same  length  as 
the  construction  lines,  but  the  same  thickness  as  the  visible  lines, 
so  as  to  distinguish  them  from  the  construction  lines. 

The  true  length  of  a  line  when  found  should  be  inked  in 
with  a  long  and  short  dash,  about  the  same  thickness  as  the 
ordinary  full  line.  When  the  true  length  of  a  line  is  given,  it 
is  put  in  like  an  ordinary  construction  line. 

Indicate  an  isolated  point  by  drawing  a  small  cross  through  it. 

37.  In  working  drawings  —  which  are  practical  applications 
of  projections  — horizontal  projections  are  usually  called  plans, 
and  vertical  projections  are  called  elevations.  Therefore,  they 
will  be  used  synonymously  throughout  this  book. 

38.  The  student  should  distinguish  between  the  terms  ver- 
tical and  perpendicular.  Vertical  is  an  absolute  term,  and 
applies  to  a  line  or  surface  at  right  angles  to  the  plane  of  the 
horizon  while  perpendicular  is  a  relative  term,  and  applies  to 
any  line  or  surface  which  is  at  right  angles  to  any  other  line  or 
surface. 

If  one  point  is  farther  from  V  than  another,  the  first  is  said 
to  be  in  front  of  the  second  point.  Hence,  if  I  say  that  a  line 
slopes  downward,  backward,  and  to  the  left,  it  signifies  that  the 
line  occupies  such  a  position  that  the  lower  end  is  nearer  V 
than  the  upper,  and  also  that  it  is  on  the  left  of  the  upper  end. 
Fig.  6  shows  the  projections  of  such  a  line. 

PROJECTIONS    OF    STRAIGHT    LINES. 

39.  A  straight  line  is  determined  by  two  points,  therefore 
it  is  only  necessary  to  draw  the  projections  of  each  end  of  a 
straight  line  and  join  them,  and  we  have  the  projections  of  the 
line.  If  the  line  is  curved,  it  becomes  necessary  to  draw  the 
projections  of  several  points  and  join  them  with  a  curve. 


70  PROJECTIONS. 

40.  If  we  lay  a  fine  wire,  ab,  Fig.  3,  on  a  horizontal  plane, 
and  also  parallel  to  a  vertical  plane,  its  horizontal  projection 
will  be  the  wire  itself,  that  is,  a^V1  is  its  horizontal  projection, 
equal  in  length  to  the  wire  itself,  parallel  to  GL,  and  at  a  dis- 
tance from  it  equal  to  the  distance  of  the  wire  from  V. 

The  vertical  projection  of  the  end  a  will  be  at  aP  in  the 
ground  line,  and  of  b  at  bv  also  in  the  ground  line,  since  each 
end  is  on  H  (Art.  9).  Hence  the  vertical  projection  of  the 
line  will  coincide  with  GL  between  av  and  bv,  and  arbv  is  its 
vertical  projection.  arbv  is  equal  in  length  to  aW,  hence  is 
equal  to  the  actual  length  of  the  wire. 

Now,  suppose  the  wire  to  be  revolved  from  right  to  left  about 
a  horizontal  axis,  through  the  end  a,  keeping  the  line  parallel 
to  V.  If  a  pencil  were  attached  to  the  end  b  at  right  angles 
to  the  line,  so  that  its  point  touched  V  at  bv,  it  would  trace  a 
circular  arc  brcvdv,  etc.  (of  which  av  is  the  centre  and  avbv,  or 
the  true  length  of  the  line,  is  the  radius)  on  the  vertical  plane 
as  the  wire  is  revolved ;  the  end  a  would,  of  course,  not  move. 
After  the  wire  has  been  revolved  through  an  angle  of  30°  its 
vertical  projection  will  be  at  avcv,  and  it  must  be  equal  in  length 
to  the  real  length  of  the  wire.  ah  will  be  the  horizontal  pro- 
jection of  the  fixed  end  of  the  wire.  Since  the  wire  is  parallel 
to  V,  every  point  in  it  is  at  the  same  distance  from  V,  hence 
their  horizontal  projections  must  all  be  the  same  distance  from 
GL,  that  is,  in  a  line  parallel  to  GL ;  the  horizontal  projec- 
tion of  the  end  c  must  also  be  in  a  line  through  c*  perpendicu- 
lar to  GL,  hence  at  ch  where  this  parallel  and  perpendicular 
intersect ;  ahch,  then,  is  the  horizontal  projection  of  the  wire 
after  it  has  been  revolved  through  an  angle  of  30°. 

For  the  same  reason  avdv  and  ahdh  are  the  two  projections 
of  the  wire  after  being  revolved  through  an  angle  of  45°. 

Similarly,  avev  and  aheh  are  its  projections  after  revolving 
through  an  angle  of  60°. 


PROJECTIONS.  71 

When  the  line  has  been  revolved  through  90°  it  becomes  per- 
pendicular to  H,  and  its  vertical  projection  is  avf",  perpendicu- 
lar to  GL,  and  its  horizontal  projection  is  a  point,  ah,  as  might 
have  been  seen  from  the  definition  of  the  projection  of  a  point. 

41.  The  same  reasoning  applies  if  we  take  the  wire  lying 
against  V  and  parallel  to  H  and  revolve  it  about  a  vertical  axis 
through  the  end  a,  as  in  Fig.  4.  The  projections  are  given  for 
the  line  lying  against  V,  and  making  angles  of  30°,  45°,  60°, 
and  90°  with  V,  being  parallel  to  H  in  each  position. 

42.  The  following  principles  may  be  noted  from  the  pre- 
ceding articles  :  — 

1st.  A  line  situated  in  either  plane  is  its  own  projection  on 
that  plane,  and  its  other  projection  is  in  the  ground  line. 

2nd.  If  a  right  line  is  perpendicular  to  either  plane  of  pro- 
jection, its  projection  on  that  plane  will  be  a  point,  and  its  pro- 
jection on  the  other  plane  will  be  perpendicular  to  the  ground 
line  and  equal  in  length  to  the  given  line. 

3rd.  When  a  line  is  parallel  to  either  coordinate  plane  its 
projection  on  that  plane  will  be  parallel  to  the  line  itself,  and 
equal  to  the  actual  length  of  the  line  in  space,  and  its  projection 
on  the  other  plane  will  be  parallel  to  the  ground  line. 

4th.  If  a  line  is  parallel  to  both  planes,  or  to  the  ground  line, 
both  projections  will  be  parallel  to  the  ground  line. 

5tk.  If  a  line  is  oblique  to  either  coordinate  plane  its  pro- 
jection on  that  plane  will  be  shorter  than  the  actual  length  of  the 
line  itself. 

6th.  If  a  line  is  parallel  to  one  coordinate  plane  and  oblique 
to  the  other,  its  projection  on  the  plane  to  which  it  is  parallel  is 
equal  to  the  true  length  of  the  line  in  space,  and  the  angle  which 
this  projection  makes  with  the  ground  line  is  equal  to  the  true 
size  of  the  angle  the  line  in  space  makes  with  the  plane  to  which 
it  is  oblique. 


72  PROJECTIONS. 

1th.  The  projection  of  a  line  on  a  plane  can  never  be  longer 
than  the  line  itself. 

8th.  If  a  point  be  on  a  line  its  projections  will  be  on  the 
projections  of  the  line. 

9th.  If  a  line  in  different  positions  makes  a  constant  angle 
with  a  plane  its  projections  on  that  plane  tvill  all  be  of  the  same 
length,  without  regard  to  the  position  it  mag  occupy  relative  to 
the  other  plane. 

43.  If  two  lines  intersect  in  space  their  projections  must  also 
intersect,  and  the  straight  line  joining  the  points  in  which  the 
projections  intersect  must  be  perpendicular  to  the  ground  line  ; 
for  the  intersection  of  two  lines  must  be  a  point  common  to 
both  lines,  whose  projections  must  be  on  the  horizontal  and  ver- 
tical projections  of  each  of  the  lines,  hence  at  their  intersec- 
tions respectively. 

44.  If  two  lines  are  parallel  in  space  their  projections  upon 
the  vertical  and  horizontal  planes  will  be  parallel  respectively. 
If  one  projection  only  of  two  lines  are  parallel,  the  lines  in 
space  are  not  parallel. 

45.  Any  two  lines  drawn  at  pleasure,  except  parallel  to  each 
other  and  perpendicular  to  the  ground  line,  will  represent  the 
projections  of  a  line  in  space. 

46.  Prob.  1.  To  draw  the  projections  of  a  line  of  a  defi- 
nite length  and  occupying  a  fixed  position  in  space. 

Let  it  be  required  to  draw  the  projections  of  a  line  1"  long, 
which  makes  an  angle  of  30°  with  H,  and  whose  horizontal 
projection  makes  an  angle  of  45°  with  GL,  the  lower  end  of 
the  line  being  £"  above  H  and  4/'  in  front  of  V. 

It  is  first  necessary  to  place  the  line  in  such  a  position  that 
its  true  length  and  the  true  size  of  the  angle  it  makes  with  one 
of  the  coordinate  planes  are  shown,  and  these  are  only  shown 


PROJECTIONS.  73 

when  it  is  parallel  to  one  of  the  coordinate  planes.  In  this  case 
it  must  be  placed  parallel  to  V.  av  and  ah,  Fig.  5,  are  the  pro- 
jections of  one  end  of  the  line,  d°  being  %"  above  GL  and  ah 
4/'  below  it.  Through  a?  draw  arb*  at  an  angle  of  30°  with 
GL  and  1"  long ;  through  ah  draw  ahb j1  parallel  to  GL,  b'1  be- 
ing found  by  dropping  a  perpendicular  from  br  (Art.  40).  The 
two  projections  of  the  line,  when  parallel  to  V,  are  thus  found 
to  be  avb*  and  ahbK 

Now  let  us  suppose  the  end  a  of  the  line  to  be  fixed  and  the 
whole  line  to  be  revolved  through  an  angle  of  45°  about  a  ver- 
tical axis  through  this  point,  the  line  keeping  the  same  angle 
with  H.  The  horizontal  projection  will  not  change  in  length 
(Art.  42,  9th),  but  will  move  through  an  angle  of  45°,  and  will 
be  found  at  ahbh.  It  is  evident  that  in  this  revolution,  so  long 
as  the  angle  with  H  does  not  change,  every  point  in  the  line  will 
remain  at  the  same  height  above  H.  The  point  a  does  not  move, 
being  in  the  axis.  We  have  seen  that  bh  moves  to  U1 ;  bv  must, 
therefore,  be  somewhere  on  a  perpendicular  through  l/\  and, 
since  the  points  do  not  change  their  heights,  it  must  also  be  on 
a  line  through  bv  parallel  to  GL,  hence  at  their  intersection  bv. 
Join  a?  and  bv  and  we  have  avb"  and  aftM  as  the  required  pro- 
jections of  the  line. 

47.  If  this  line  were  revolved  through  15°  more,  the  point 
b11  would  go  to  ch,  and  bD  to  cv,  and  a?c°  and  ahch  would  be  the 
projections  of  the  line  making  an  angle  of  30°  with  H,  and 
whose  horizontal  projection  made  an  angle  of  60°  with  GL. 

If  it  were  revolved  still  30°  more,  the  two  projections  would 
be  avd"  and  ahdh,  each  being  perpendicular  to  the  ground  line. 
When  a  line  is  in  this  position,  i.  e.,  has  its  two  projections  in 
a  Hue  perpendicular  to  GL,  it  is  said  to  be  in  a  profile  plane, 
a  profile  plane  being  understood  to  be  one  that  is  perpendicular 
to  both  V  and  H. 


74  PROJECTIONS. 

When  a  line  is  oblique  to  only  one  of  the  coordinate  planes 
it  is  said  to  make  a  simple  angle ;  when  it  is  oblique  to  both  of 
them  it  is  said  to  make  a  compound  angle. 

48.  If  the  angle  that  the  line  made  with  V  had  been  given, 
it  would  have  been  necessary  to  have  first  placed  the  line  par- 
allel to  H,  and  then  to  have  revolved  it  about  an  axis  through 
one  end  perpendicular  to  V,  in  which  case  the  length  of  the 
vertical  projection  would  not  change,  and  the  points  would  not 
change  their  distances  from  V. 

In  Fig.  6,  avbv  and  ahbh  are  the  two  projections  of  a  line  1" 
long,  making  an  angle  of  45°  with  V,  and  whose  vertical  pro- 
jection makes  an  angle  of  60°  with  GL.  The  principles  and 
explanation  for  this  construction  are  the  same  as  for  Prob.  1, 
if  the  horizontal  and  vertical  planes  are  supposed  to  be  inter- 
changed. 

49.  Prob.  2.  To  find  the  true  length  of  a  line  given  by  its 
projections,  and  the  angle  it  makes  with  either  plane  of  projec- 
tion. 

Let  avbv  and  ahbh,  Fig.  7,  be  the  projections  of  the  given  line. 
The  true  length  is  only  shown  when  it  is  parallel  to  one  of  the 
coordinate  planes,  hence  this  line  must  be  revolved  about  an 
axis  through  either  end  until  it  is  parallel  to  one  of  the  planes. 
If  it  is  revolved  about  a  vertical  axis  through  a  until  it  is  par- 
allel to  V,  the  point  a  does  not  move,  bh  moves  to  bh,  bv  is  found 
at  6"  (where  a  perpendicular  through  bh  intersects  a  horizontal 
through  bv),  and  avhj  is  the  true  length.  Also,  the  angle  which 
a"bv  makes  with  GL  is  equal  to  the  true  size  of  the  angle  the 
line  makes  with  H. 

If  it  had  been  required  to  find  the  angle  this  line  made  with 
V,  it  would  have  been  necessary  to  have  revolved  the  line  about 
a  horizontal  axis  until  it  became  parallel  to  H.     Assuming  the 


PROJECTIONS.  75 

axis  through  the  end  b,  Fig.  7,  dP  moves  to  a *,  ah  to  ah,  and 
aA5ft  is  the  true  length  of  the  line  (which  of  course  should  equal 
avb"),  and  the  angle  it  makes  with  GL  is  equal  to  the  true  size 
of  the  angle  the  line  makes  with  V. 

50.  Note.  The  angles  which  the  vertical  and  horizontal 
projections  of  a  line  make  with  GL  are  greater  than  the  angles 
which  the  line  in  space  males  with  H  and  V  respectively,  except 
when  the  line  is  parallel  to  one  of  the  planes. 

PROJECTIONS    OF    SURFACES. 

51.  Plane  surfaces  are  bounded  by  lines,  therefore  the  prin- 
ciples which  govern  the  projections  of  lines  are  equally  appli- 
cable to  these  surfaces. 

52.  If  we  suppose  a  rectangular  card  abed,  Fig.  8,  placed 
with  its  surface  parallel  to  V  and  perpendicular  to  H,  each  edge 
being  parallel  to  V,  it  will  be  projected  on  V  in  a  line  equal  and 
parallel  to  itself,  hence  the  true  size  of  the  card  itself  is  shown 
in  vertical  projection.  Two  of  the  edges,  ab  and  cd,  being  per- 
pendicular to  H,  are  projected  on  that  plane  in  the  points  ah 
and  dh  respectively.  The  other  two  edges,  ad  and  be,  are  par- 
allel to  H  as  well  as  V,  hence  they  will  be  projected  in  their 
true  length  on  H.  and,  since  one  is  vertically  over  the  other, 
they  will  both  be  horizontally  projected  in  the  same  line  ahdh. 

Now,  if  the  card  is  revolved  about  one  of  its  vertical  edges 
as  an  axis,  like  a  door  on  its  hinges,  the  vertical  edge  which 
coincides  with  the  axis  does  not  move ;  the  other  vertical  edge 
moves  in  the  arc  of  a  circle.  The  horizontal  projection  of  the 
card  will  still  be  a  straight  line  of  the  same  length  as  before. 
Let  the  card  be  revolved  through  60°  ;  ah,  does  not  move  ;  dh 
moves  in  the  arc  of  a  circle,  of  which  ah  is  the  centre  and  ahdh 
the  radius,  to  dh ;  ahdh  is  the  horizontal  projection  of  the  card 
in   its  new  position  ;  the  vertical  projection  of  the  edge  cd  in 


7G  PROJECTIONS. 

this  jjosition  is  found  at  ctvdtv,  vertically  above  dh,  and  arbvc'dv 
is  the  vertical  projection  of  the  card  after  being  revolved 
through  an  angle  of  60°.    . 

If  the  card  should  be  revolved  through  30°  more,  i.  e.,  90° 
in  all,  its  surface  will  be  at  right  angles  with  both  coordinate 
planes,  and  its  two  projections  will  be  found  at  avbv  and  aV',  in 
one  and  the  same  straight  line  perpendicular  to  GL. 

53.  If  the  card  be  placed  on  H,  with  one  of  its  edges  par- 
allel to  V,  ahbhchdh,  Fig.  9,  will  be  its  horizontal  and  arbv  its 
vertical  projection.  If  this  card  be  revolved  about  one  of  its 
edges  which  are  perpendicular  to  V  as  an  axis,  like  a  trap-door 
on  its  hinges,  through  an  angle  of  30°,  arbv  and  ahbihcihdh  will 
be  its  two  projections.  If  it  be  revolved  through  60°  more,  or 
90°  in  ail,  its  projections  will  be  avev  and  ahdh,  which  are  just 
the  same  as  avbv  and  aheh  in  Fig.  8,  as  they  should  be,  since  the 
cards  are  the  same  size  in  the  two  figures  and  they  occupy  the 
same  relative  position  in  each,  i.  e.,  they  are  in  a  profile  plane. 

54.  The  following  principles  may  be  noted  :  — 

1st.  When  a  plane  surface  is  perpendicular  to  another  plane 
its  projection  on  that  plane  will  be  a  line. 

2nd.  When  a  plane  surface  is  parallel  to  either  coordinate 
plane,  its  projection  on  that  plane,  will  be  equal  to  the  true  size 
of  the  surface  and  its  other  projection  will  be  a  line  parallel  to 
GL. 

3rd.  When  a  plane  surface  is  perpendicular  to  one  plane  and 
oblique  to  the  other,  the  angle  which  its  projection  on  the  plane  to 
which  it  is  perpendicular  makes  with  the  ground  line  is  equal  to 
the  angle  the  surface  in  space  makes  with  the  plane  to  which  it  is 
oblique. 

4th.  If  a  plane  surface,  in  different  position*,  makes  a  con- 
stant angle  with  a  plane,  its  projections  on  that  plane  will  all  be 
of  the  same  size. 


I 
PROJECTIONS.  77 

55.  Prob.  3.  To  draw  the  two  projections  of  a  plane  sur- 
face, or  card,  of  a  certain  size,  and  making  a  compound  angle 
with  the  coordinate  planes. 

Let  the  card  be  of  the  size  shown  in  Fig.  9,  and  suppose  it 
to  make  an  angle  of  30°  with  H,  and  its  horizontal  projection 
an  angle  of  45°  with  GL. 

Draw  the  horizontal  projection  allbhc!ld/l  of  the  card  equal  to 
its  true  size ;  avb''  will  be  its  vertical  projection.  Revolve  arb° 
through  an  angle  of  30°  to  avbv,  and  avbv  will  be  the  vertical 
projection  of  the  card  when  it  makes  an  angle  of  30°  with  H 
and  is  perpendicular  to  V  ;  al'b^c^d'1  is  its  corresponding  hori- 
zontal projection. 

The  angle  with  H  is  still  to  be  30°  after  the  card  has  been 
revolved  to  its  desired  position,  hence  its  horizontal  projection 
will  be  the  same  size.  Therefore,  make  ahb]lc}ldh,  Fig.  10, 
equal  in  size  to  ahb]'c]'dh,  Fig.  9,  and  making  the  desired  angle 
with  GL.  In  this  revolution,  as  long  as  one  edge  rests  on  H 
and  the  angle  remains  constant  with  H,  every  point  keeps  the 
same  height  above  IT,  therefore  the  vertical  projections  of  ah 
and  dh,  Fig.  10,  must  be  found  at  av  and  dv  ;  also  of  b]1  and  cA 
at  6"  and  c ",  whose  heights  above  GL  are  equal  to  the  height 
of  b;\  Fig.  9,  above  GL. 

The  other  parallelograms  in  Fig.  10  represent  the  projections 
of  the  same  card  at  different  angles  with  H,  the  horizontal  pro- 
jections making  the  same  angles  with  GL. 

56.  ahb''chdh  and  avdvbvcv,  Fig.  11,  represent  the  projections 
of  the  same  card  when  it  is  lying  on  H  with  one  of  its  diago- 
nals parallel  to  V,  and  a''d''b'c"  and  a,lb//ic''dh  are  its  projec- 
tions after  being  revolved  about  an  axis  perpendicular  to  V 
through  the  corner  a  through  an  angle  of  45°.  Fig.  12  repre- 
sents the  projections  of  this  card  when,  besides  making  an  angle 
of  45°  with  H,  the  horizontal  projection  of  the  diagonal  makes 


78  PROJECTIONS. 

an  angle  of  30°  with  GL.  The  steps  to  obtain  this  are  exactly 
the  same  as  in  Figs.  9  and  10,  hence  the  explanation  will  not 
be  repeated. 

57.  Prob.  4.  To  draw  the  projections  of  a  regular  pentag- 
onal card,  the  diameter  of  the  circumscribed  circle  being  given, 
in  two  positions.  1  st ,  when  it  is  perpendicular  to  V  and  making 
an  angle  of  60°  with  H,  one  of  its  edges  being  perpendicular  to 
V ;  2nd,  when,  besides  making  an  angle  of  60°  with  H  as  in 
1st  position,  it  has  been  revolved  through  an  angle  of  45°. 

The  pentagon  must  first  be  drawn  in  its  true  size  and  posi- 
tion ;  a^J/'c/'d^e*,  Fig.  13,  equal  to  the  actual  size  of  the  card, 
is  its  horizontal  projection,  cj'd'1  being  perpendicular  to  GL, 
and  arbrcv  is  its  vertical  projection.  Revolve  a'b^'c1'  through 
an  angle  of  60°  to  arbvcv ;  each  point  moves  in  the  arc  of  a  cir- 
cle with  a  as  a  centre,  and  arlfcv  will  be  the  vertical  projection 
of  the  card  when  in  the  1st  position  asked  for  ;  ahbhchdheh  is 
its  corresponding  horizontal  projection. 

For  the  2nd  position  revolve  the  plan  just  found  through  45° 
to  the  position  ahbh,  etc.,  shown  in  Fig.  14;  arb''c'drev  will  be 
the  corresponding  vertical  projection. 

58.  Cards  of  any  shape  and  size,  and  occupying  any  position, 
may  be  drawn  in  the  same  way,  care  being  taken  to  locate  one 
point  at  a  time. 

59.  If  the  angle  the  card  made  with  V  had  been  given,  it 
would  have  been  necessary  to  have  first  placed  the  card  par- 
allel to  V  and  then  to  have  revolved  it,  through  the  angle  it 
made  with  V,  about  an  axis  perpendicular  to  II,  in  which  case 
the  length  of  the  horizontal  projection,  which  is  a  straight  line, 
would  not  change,  and  the  several  points  would  not  change 
their  respective  distances  from  H. 


PROJECTIONS.  79 

In  the  second  revolution,  which  chauges  the  angle  with  the 
coordinate  planes  from  simple  to  compound,  the  vertical  projec- 
tion must  be  revolved  and  the  corresponding  plan  found.  The 
angle  with  V  being  the  same  the  vertical  projection  does  not 
change  its  size  ;  the  distances  of  the  points  in  front  of  V  remain 
the  same  after  revolution  as  before,  hence  are  found  at  the  same 
distances  from  GL  respectively. 

60.  Prob.  5.  To  draw  the  projections  of  a  circular  card 
making  a  compound  angle  with  the  coordinate  planes. 

Let  the  diameter  of  the  card  be  given,  the  angle  it  makes 
with  V,  and  the  angle  through  which  the  vertical  projection  is 
to  be  revolved. 

A  circle  may  be  considered  as  a  polygon  of  an  infinite  num- 
ber of  sides,  hence  we  can  take  as  many  points  as  we  please  on 
the  circumference  of  the  circle,  and  each  one  moves  according 
to  the  principles  just  described. 

Place  the  card  parallel  to  V ;  a  circle,  arb''cf,  etc.,  Fig.  15, 
equal  to  the  actual  size  of  the  given  circle,  is  its  vertical  pro- 
jection, and  ahb''c\  etc.  is  its  horizontal  projection. 

Revolve  the  card  through  the  required  angle  about  a  vertical 
axis  through  a;  a,lbl'ch,  etc.  is  its  horizontal,  and  avbvc",  etc.  is 
its  vertical  projection. 

The  card  is  to  be  revolved  through  a  certain  angle,  still 
keeping  the  same  angle  with  V.  The  size  of  the  vertical  pro- 
jection will,  therefore,  not  change.  Hence,  revolve  the  vertical 
projection  found  in  Fig.  15  through  the  required  angle  to  the 
position  acb'c'\  etc.,  Fig.  16.  None  of  the  points  change  their 
distance  from  V,  consequently  a''bhch,  etc.  is  the  horizontal  pro- 
jection of  the  card,  found  as  in  the  last  problem. 

61.  Fig.  17  shows  a  somewhat  shorter  method  of  drawing 
the  projections  of  a  circular  card  making  a  simple  angle  with 


80  PROJECTIONS. 

the  coordinate  planes ;  in  this  case  it  makes  an  angle  of  30° 
with  V  and  is  perpendicular  to  II.  ahbhc]\  etc.,  making  30° 
with  GL,  is  its  horizontal  projection.  Suppose  the  card  re- 
volved about  its  horizontal  diameter  ae  until  it  is  parallel  to  H. 
It  will  then  be  shown  in  its  true  size  at  ahbc,  etc. ;  bV1,  cch, 
etc.  will  show  the  actual  distances  of  the  points  b,  c,  etc.  from 
the  horizontal  diameter ;  cfev  will  represent  the  vertical  projec- 
tion of  the  horizontal  diameter  about  which  the  card  is  revolved. 
Of  course,  bv  must  be  found  somewhere  in  a  line  through  bh, 
perpendicular  to  GL,  therefore,  lay  off  tbv  equal  to  bf/',  and  b* 
is  a  point  of  the  required  vertical  projection  ;  c''s  is  made  equal 
to  cc\  d'r  to  ddh,  etc.  Other  points  may  be  found  in  the  same 
way. 

It  is  evident  that  n  V\  meh,  etc.  are  respectively  equal  to 
bbh,  ccb,  etc.,  hence  it  is  only  necessary  to  revolve  the  semi- 
circle and  the  distance  bbh  is  laid  off  on  both  sides  of  the  diam- 
eter are^  giving  the  two  points  b1'  and  ri". 

PROJECTIONS    OF    SOLIDS. 

62.  A  cube  is  a  solid  bounded  by  six  equal  faces,  and  when 
it  is  placed  so  that  two  of  its  faces  are  parallel  to  H.  and  two 
others  parallel  to  V,  its  two  projections  are  a'b'e'f'  and  ahbhchdh, 
Fig.  18.  The  top  and  bottom,  being  parallel  to  H,  are  hori- 
zontally projected  in  one  and  the  same  square,  ct!'V'chdh,  which 
is,  of  course,  equal  to  the  exact  size  of  these  faces,  and  their 
vertical  projections  are  av&°  and  evf"  respectively  (Art.  30-2nd)  ; 
the  front  and  back  faces  being  parallel  to  V  are  vertically  pro- 
jected in  one  and  the  same  square,  arb''crf'\  which  is  also  equal 
to  the  exact  size  of  these  faces,  and  their  horizontal  projections 
are  ahbh  and  chdh  respectively  (Art.  30-2nd)  ;  the  left  and  right 
hand  faces  are  perpendicular  to  both  V  and  H,  therefore  their 
vertical  projections  are  avev  and  ///'',  and  their  horizontal  pro- 
jections are  ahdh  and  bhch  respectively  (Art.  30-lst). 


PROJECTIONS.  81 

The  plan  shows  two  dimensions  of  the  cube,  the  length  and 
breadth,  and  the  elevation  two  dimensions,  the  length  and  thick- 
ness ;  therefore,  the  three  dimensions  of  the  solid  being  shown 
in  their  true  size  in  the  two  projections  the  object  is  com- 
pletely represented.  In  this  case  it  does  not  matter  which  pro- 
jection is  drawn  first,  as  they  each  show  two  dimensions  in  their 
true  size. 

63.  Shade  Lines.  In  outline  drawings  it  is  customary  to 
put  in  shade  lines,  i.  e.,  lines  heavier  than  the  others  ;  they  give 
relief  to  the  drawing,  and,  when  properly  placed,  are  of  assist- 
ance in  reading  it. 

Shade  lines,  or  edges,  are  those  edges  which  separate  light, 
from  dark  surfaces. 

The  rays  of  light  are  generally  assumed  to  come  from  over 
the  left  shoulder  in  the  direction  of  the  diagonal  of  a  cube,  the 
person  supposed  to  be  facing  the  cube,  and  the  cube  to  be  in 
the  position  shown  in  Fig.  18.  That  is,  the  ray  of  light  enters 
the  cube  at  the  upper,  front,  left-hand  corner,  whose  projections 
are  a?  and  ah,  and  leaves  it  at  the  lower,  back,  right-hand  corner, 
whose  projections  are  f9  and  ch ;  the  diagonal  joining  these 
points  will  represent  the  actual  direction  of  the  conventional 
ray  of  light,  and  its  projections  avf"  and  ahch  are  the  projec- 
tions of  this  ray.  The  different  rays  of  light  are  all  supposed 
to  be  parallel  to  each  other. 

It  is  evident  from  the  figure  that  both  projections  of  the  rays 
of  light  make  angles  of  45°  with  GL.  The  student  should 
distinctly  understand  that,  although  the  projections  of  the  ray 
of  light  make  45°  with  GL,  the  actual  angle  it  makes  with  V 
or  H  is  quite  different.  To  find  this  angle  apply  the  princi- 
ples of  Art.  25  to  Fig.  18,  and  we  get  a=  35°  15'  52"  as  its 
actual  size. 

C>1.     In  the  cube,  Fig.  18,  the  top,  front,  and  left-hand  faces 


82  PROJECTIONS. 

are  light,  and  the  bottom,  back,  and  right-hand  faces  are  dark, 
and  the  shade  lines  are,  therefore,  erf'\  which  separates  the 
front  from  the  bottom,  b''f'\  which  separates  the  front  from  the 
right-hand  face,  buc'\  which  separates  the  top  from  the  right- 
hand  face,  and  chd'\  which  separates  the  top  from  the  back  face. 
The  two  other  shade  edges  which  separate  the  left-hand  face 
from  the  back  and  bottom  faces  are  not  seen  in  either  projec- 
tion, since  the  top  and  front  edges  of  the  left-hand  face  are  in 
the  same  plane  and  nearer  the  eye. 

It  is  for  this  same  reason  that  the  shade,  lines  mentioned 
above  are  seen  only  in  one  projection,  i.  e.,  erfv  is  a  shade  line; 
it  is  seen  in  elevation,  but  in  plan  is  hidden  by  the  upper  front 
edge  of  the  cube  avbv-ahLh. 

65.  It  will  be  noticed  that  the  right-hand  and  lower  edges 
are  shaded  in  elevation,  and  the  right  hand  and  upper  in  plan. 
From  this  many  draftsmen  have  adopted  the  arbitrary  rule  to 
shade  the  right-hand  and  lower  lines  in  elevation,  and  the  right- 
hand  and  upper  in  plan.  This  rule  is  really  applicable  in  but 
few  cases,  except  when  the  object  is  of  rectangular  section  and 
so  placed  that  its  surfaces  are  perpendicular  to  one  or  both  of 
the  coordinate  planes. 

Other  draftsmen  shade  the  right-hand  and  lower  lines  in 
both  plan  and  elevation.  This  is  also  applicable  only  in  the 
cases  stated  above,  besides  being  obliged  to  change  the  direction 
of  the  ray  of  light  in  plan  and  elevation,  or  imagine  the  object 
revolved  while  the  ray  of  light  remains  fixed. 

Others  still^ollovv  the  rule  given  in  Art.  63,  except  that  they 
only  put  in  shade  lines  where  the  dark  portion  of  space  adja- 
cent to  the  line  in  question  is  visible,  and  they  change  the  direc- 
tion of  the  ray  of  light,  or,  what  is  practically  the  same  thing. 
revolve  the  object.  By  this  method  the  plan  and  elevation  are 
shaded  alike,  but  the  right-hand  line  is  not  necessarily  a  shade 
line,  while  the  left-hand  line  is  necessarily  not  a  shade  line. 


PROJECTIONS.  83 

The  method  taken  up  in  this  book,  and  the  one  it  is  expected 
that  the  student  will  follow  in  this  course,  is  not  given  because 
it  is  the  one  most  generally  in  use,  or  because  it  is  the  easiest, — 
quite  the  contrary ;  but  because  it  is,  in  the  opinion  of  the 
writer,  the  only  method  which  can  be  followed  consistently 
throughout  a  course  of  projections,  shadows,  isometric,  and 
working  drawings. 

66.    Prob.  6.     To  draw  the  two  projections  of a  right  square 

prism  with  its  base  on  II  and  its  vertical  faces  oblique  to  V. 
Fig.  19. 

It  is  evident  that  the  vertical  faces  will  not  be  projected  on 
V  in  their  true  size,  the  height  only  of  the  prism  being  shown 
in  its  real  length,  and  also  that  the  two  ends,  being  parallel  to 
H,  will  be  projected  on  that  plane  in  their  true  size.  Hence, 
draw  the  square  ahbhchdh  equal  to  the  ends  of  the  prism,  with 
its  edges  making  the  same  angles  with  GL  as  the  vertical  faces 
make  in  space  with  V.  Through  the  corners  ah,  bh,  ch,  and  dh, 
draw  the  perpendiculars  avev,  b'fv,  etc.,  making  each  equal  in 
length  to  the  height  of  the  prism.  avbrcrd"  will  be  the  vertical 
projection  of  the  upper  base.  evfvmvnv  of  the  lower  base,  and 
avc?mvev  the  vertical  projection  of  the  whole  prism. 

In  this  case  the  top  and  two  front  faces  are  light,  while  the 
bottom  and  two  back  faces  are  dark  ;  hence,  in  plan  the  lines 
ahdh  and  chdh,  which  separate  the  top  from  the  two  back  faces, 
-are  shade  lines  ;  in  elevation  they  are  behind  the  two  front  edges 
ab  and  be,  consequently  are  not  seen.  The  element  ae  separates 
the  left  front  from  the  left  back  face,  hence  is  a  shade  line,  and 
ari''\  its  vertical  projection,  is  accordingly  made  heavy  ;  its  hori- 
zontal projection  is  simply  a  point;  the  element  cm  also  sepa- 
rates a  light  from  a  dark  surface,  and  its  visible  projection  c'ni0 
should  be  made  heavy.     The  edges  ef  and  fm  separate  the  two 


84  PROJECTIONS. 

front  faces  from  the  bottom,  and  their  visible  projections  erfv 
and  /'in''  are  also  made  heavy. 

67.  Prob.  7.  To  draw  the  two  projections  of  a  right  regu- 
lar •pentagonal  prism  standing  with  -its  base  on  H,  with  none  of 
its  faces  parallel  to  V.     Fig.  20. 

Here,  as  in  the  last  problem,  it  is  necessary  to  draw  the  hori- 
zontal projection  first,  as  it  shows  the  pentagonal  end  in  its  true 
size  and  position.  ahV'chdheh  is  its  horizontal  projection,  and 
avdvri°fv,  found  as  in  the  last  problem,  is  its  vertical  projection. 

From  the  shade  lines  shown  in  the  figure  it  will  be  noticed 
that  the  line  cd,  which  separates  the  light  top  from  the  dark 
right-hand  face,  is  visible  in  both  projections,  hence  it  is  made 
heavy  in  both. 

A  prism  of  any  number  of  sides  standing  on  II  can  be  drawn 
in  the  same  manner. 

68.  Prob.  8.  To  draw  the  two  pro  ect ions  of  a  right  regu- 
lar hexagonal  prism  with  its  axis  perpendicular  to  V.     Fig.  21. 

Here  it  is  necessary  to  draw  the  vertical  projection  first,  and 
construct  the  horizontal  projection  from  it  according  to  the 
principles  noted  in  the  last  two  problems. 

60.  From  an  inspection  of  Figures  18  to  21  it  is  evident  that 
the  4.VJ  triangle  can  be  used  to  determine  positiccly  the  light  and 
dark  fares  only  when  these  faces  are  perpendicular,  or  nearly 
so,  to  one  or  both  of  the  coordinate  planes. 

In  Fig.  18  the  triangle  can  be  used  iu  both  plan  and  eleva- 
tion, since  every  face  is  perpendicular  to  at  least  one  of  the 
coordinate  planes.  In  Figs.  19  and  20  the  faces  are  perpen- 
dicular to  H  only  (except  the  top,  which  is,  of  cuurse,  known  to 
be  light),  hence  the  45°  triangle  can  only  be  used  iu  the  plan. 


#.  &  vh  y £a* 


PROJECTIONS.  80 

In  Fig.  21  the  faces  are  perpendicular  to  V  only  (except  the 
front  end,  which  must  be  light),  hence  the  triangle  can  only  be 
used  in  elevation. 

70.  Prob.  9.  To  draw  the  two  projections  of  a  right  cylin- 
der standing  on  its  base.     Fig.  22. 

Its  horizontal  projection  will  be  a  circle  equal  to  the  end  of 
the  cylinder  (Art.  42).  Its  vertical  projection  will  be  repre- 
sented by  avbrdvcv,  ar<f  being  the  vertical  projection  of  the  top, 
bvdv  of  the  bottom,  avbr  and  cvdv  of  the  extreme  left  and  right 
hand  elements,  or  the  contour  lines  as  they  are  called. 

By  applying  the  45°  triangle  it  is  evident  that  the  shade  line 
in  plan  will  be  the  half  circle  nihc,'oh  between  the  points  where 
the  triangle  is  tangent  to  the  circle.  The  element  cd  is  not  a 
shade  line,  as  it  does  not  separate  a  light  from  a  dark  surface. 
The  shade  element  would  be  or,  but  as  it  is  not  drawn  of  course 
it  cannot  lie  shaded. 

The  bottom  of  the  cylinder  is  dark,  and  strictly  the  line  b''rv 
would  be  heavy,  leaving  the  portion  r''d1'  as  light ;  but,  since  it 
is  practically  impossible  to  stop  the  shade  line  at  the  point  r° 
and  make  a  good-looking  line,  I  should  disregard  this  short  piece 
aud  shade  the  line  the  whole  length.  Similarly,  on  top  there 
will  be  a  short  piece  between  ov  and  cv  that  would  strictly  be 
shaded,  but  for  the  same  reason  I  would  disregard  this  and 
make  the  whole  top  a  light  line. 

The  absence  of  the  shade  line  cvdv  in  the  vertical  projection 
enables  us  to  tell  at  once  that  the  object  is  cylindrical  even 
before  we  look  at  its  plan. 

A  cone  is  like  a  cylinder,  except  that  its  elements  all  intersect 
in  a  common  point  called  the  vertex,  while  in  the  cylinder  they 
are  all  parallel.    Fig.  51  represents  the  two  projections  of  a  cone. 


86  PROJECTIONS. 

71.  Prob.  10.  To  draw  the  two  projections  of  a  right  regu- 
lar hexagonal  pyramid  with  its  base  resting  on  H.      Fig.  23. 

As  we  look  directly  down  upon  a  pyramid,  in  this  position 
we  shall  see  all  of  its  sloping  faces,  and  consequently  the  edges 
which  separate  these  faces.  We  shall  also  see  the  edges  which 
separate  these  sloping  faces  from  the  base,  i.  e.,  the  outline 
of  the  base.  In  this  case  the  base  is  parallel  to  H,  hence  its 
outline  is  shown  in  plan  equal  to  the  real  size  of  the  base. 
Therefore,  the  regular  hexagon,  ahbhchdhehfh,  is  the  horizontal 
projection  of  the  outline  of  the  base.  Now,  since  in  a  right 
pyramid  the  base  is  perpendicular  to  the  axis,  it  will  be  easily 
seen  that  the  horizontal  projection  of  the  vertex  of  the  pyramid 
must  be  at  oh,  the  centre  of  the  hexagon. 

Drawing  lines  from  oh  to  each  corner  of  the  hexagon,  the 
horizontal  projection  of  the  pyramid  is  completed.  The  points 
in  the  base  are,  of  course,  vertically  projected  in  GL,  the  ver- 
tex at  ov  at  a  distance  above  GL  equal  to  the  altitude  of  the 
pyramid.  Joining  ov  with  av,bv,ct',  etc.  we  have  its  vertical 
projection. 

The  shade  lines  of  a  pyramid  are  not  found  directly  by  means 
of  the  45°  triangle,  as  we  have  been  able  to  do  previous  to  this, 
on  account  of  the  faces  not  being  perpendicular  to  either  coor- 
dinate plane.  If  we  try  to  use  the  triangle  as  in  the  case  of 
the  prism,  we  would  have  said  that  the  three  faces,  /''o^a1', 
a!'ohbh,  and  bhohch  were  light,  and  the  three  remaining  faces 
dark,  but  this  is  not  the  case.  For  let  us  suppose  that  the  alti- 
tude of  this  pyramid  is  so  small  that  each  of  the  faces  of  the 
pyramid  makes  an  angle  with  H  less  than  35°  16'  (the  angle 
the  ray  of  light  makes  with  H).  It  is  evident  that  all  of  the 
sloping  faces  will  be  light,  and  the  bottom  being  dark  the  shade 
lines  would  go  entirely  round  the  base.  Now,  if  we  consider 
the  altitude  to  increase,  we  shall  soon  reach  the  point  when  the 

N 


PROJECTIONS.  87 

face  ohdheh  will  become  dark,  all  of  the  rest  remaining  light, 
and  the  shade'line  would  change  from  dheh  to  ehoh  and  ohdh.  If 
the  altitude  be  still  further  increased,  we  next  get  the  case  shown 
in  the  h'gure  where  the  ia,c,Q  fhoheh  becomes  dark,  and  the  shade 
lines  would  change  from  ehfh  and  ehoh  to  fhoh.  If  the  altitude 
should  be  still  further  increased,  the  face  chohdh  would  presently- 
become  dark  also. 

Of  course  the  other  three  faces  would  never  become  dark 
while  the  pyramid  rested  on  its  base,  even  if  the  vertex  were 
extended  to  infinity,  in  which  case  we  should  simply  have  a 
prism.  In  cases  like  this,  or  where  any  surface  is  oblique  to 
both  V  and  II,  it  is  necessary  to  find  the  shadow  of  the  object, 
thus  determining  which  surfaces  are  light  and  which  are  dark. 

72.  In  Fig.  24  arbrfre"  is  the  elevation  and  ahbhchdk  is  the 
plan  of  a  rectangular  prism,  with  two  of  its  faces  parallel  to  each 
of  the  coordinate  planes.  The  plan  shows  its  length  and  width, 
and  the  elevation  its  length  and  thickness.  If  a  side  elevation 
is  desired,  it  will  show  the  width  and  thickness.  To  get  this 
the  object  must  be  projected  onto  a  plane  at  right  angles  to  the 
two  coordinate  planes,  i.  e.,  the  profile  plane,  and  this  plane 
revolved  about  its  intersection  with  V,  as  an  axis,  to  coincide 
with  V.  POR  is  such  a  plane,  resting  against  the  end  of  the 
prism.  PO  being  its  intersection  with  V  and  OR  its  intersection 
with  H. 

In  this  revolution  none  of  the  points  change  their  heights 
above  H.  nor  their  distances  from  the  axis  PO,  hence  the  rect- 
angle bjc^f^mj  will  represent  this  side  elevation,  it  being  of 
course  the  same  height  above  OL  that  the  front  elevation  is, 
and  the  distance  that  c "/»"  is  from  the  axis  PO  will  be  equal  to 
the  distance  the  back  of  the  prism  is  in  front  of  V. 

The  shade  lines  in  the  end  elevation  are  shaded  the  same  way 


88  PROJECTIONS. 

as  in  the  front  elevation  ;  the  ray  of  light  is  supposed  to  come 
from  over  the  person's  left  shoulder  when  he  is  facing  the  pro- 
file plane,  i.  e.,  the  vertical  projection  of  the  ray  of  light  is  the 
same  for  all  elevations. 

73.  Pkob.  11.  To  draw  the  plan  and  two  elevations  of  a 
square  prism  with  its  axis  parallel  to  and  at  a  definite  distance 
from  both  V  and  H,  all  of  its  faces  being  oblique  to  both  V  and 
H.     Fig.  25. 

The  end  elevation  is  the  only  view  of  the  prism  which  shows 
one  of  its  surfaces  in  its  true  size  and  position  relative  to  the 
coordinate  planes,  hence  this  view  must  be  drawn  first. 

Locate  the  point  ov  at  a  perpendicular  distance  above  GL 
equal  to  the  height  of  the  axis  above  H ;  through  ov  draw  two 
lines  at  right  angles  to  each  other,  making  angles  with  GL  equal 
to  those  made  by  the  long  faces  of  the  prism  with  H  respect- 
ively ;  lay  off  on  each  of  these  lines,  on  both  sides  of  ov,  a  dis- 
tance equal  to  half  the  side  of  the  square ;  through  these  points 
draw  lines  parallel  to  the  lines  through  o1',  and  the  square  a''b-c* 
dv  thus  formed  will  be  the  end  elevation  of  the  prism  in  its  cor- 
rect position. 

To  locate  the  axis  the  correct  distance  from  V  draw  POR, 
which  represents  the  profile  plane,  perpendicular  to  GL  and  at 
a  horizontal  distance  to  the  right  of  ov,  equal  to  the  distance  of 
the  axis  in  front  of  V. 

In  the  last  problem  the  end  view  was  constructed  from  the 
plan  and  front  elevation  ;  in  this  problem  we  construct  the 
plan  and  front  elevation  from  the  end  view  by  simply  reversing 
the  steps. 

The  horizontal  lines,  arev,  bvf\  cl'mv,  and  dvnv,  drawn  through 
av,  bv,  cv,  and  dv  respectively,  and  each  equal  in  length  to  the 
length  of  the  prism,  will  be  the  front  elevation  of  the  different 


PROJECTIONS.  89 

elements  of  the  prism ;  joining  these  ends  the  front  elevation 
of  the  prism  is  complete. 

From  O  along  OR  lay  off  Omh,  Of1,  etc.  equal  to  the  hori- 
zontal distances  of  cj,  #",  etc.  from  PO.  Through  these  points 
mh,f,  etc.  draw  the  horizontal  lines  mhch,  fV1,  etc.,  each  equal 
in  length  to  the  length  of  the  prism,  and  joining  the  ends  the 
plan  of  the  prism,  ahchmheh,  will  be  complete. 

The  long  faces  of  the  prism  being  perpendicular  to  V  in  the 
end  view,  the  shade  lines  for  that  view  may  be  found  directly 
by  using  the  45°  triangle,  as  shown  by  the  arrows.  In  revolv- 
ing the  prism  from  the  position  shown  in  end  view  to  that  in 
the  front  view,  the  front  and  back  ends  change  from  light  to 
dark  and  from  dark  to  light  respectively,  but  the  long  faces  are 
light  or  dark  in  the  front  view  and  plan  according  as  they  are 
light  or  dark  in  the  end  view  (provided  the  projection  of  the 
right-hand  end  is  represented,  which  will  be  seen  to  the  left  of 
the  front  view). 

74.  Prob.  12.  To  draw  the  two  projections  of  a  regular 
pentagonal  prism,  with  its  axis  parallel  to  H  and  oblique  to  V, 
and  its  lower  left-hand  long  face  making  a  definite  angle  with  H. 
Fig.  26. 

Here,  as  in  the  last  problem,  it  is  necessary  to  draw  the  view 
of  the  end  of  the  prism  when  its  axis  is  perpendicular  to  V,  so 
as  to  show  it  in  its  true  size  and  position,  a^b^c^d^e^  is  its 
end  view,  the  edge  a "£"  making  an  angle  with  GL  equal  to  the 
angle  the  lower  left  face  makes  with  H.  In  the  last  problem 
the  prism  was  revolved  through  an  angle  of  90°  to  its  actual 
position,  but  in  this  it  is  revolved  through  a  smaller  angle. 
The  steps  being  otherwise  just  the  same,  the  explanation  will 
not  be  repeated. 

The  shade  lines  in  this  case  may  also  be  found,  as  in  the  last 
problem,  by  using  the  45°  triangle  on  the  end  view. 


90  PROJECTIONS. 

75.  Let  us  now  suppose  a  case  where  the  edge  c*d*,  Fig. 
26,  makes  an  angle  of  40°  with  H  and  in  the  same  direction. 
It  is  evident  that  when  the  axis  of  the  prism  is  perpendicular 
to  V  the  surface  which  is  projected  in  cvdv  will  be  light.  Now, 
if  the  prism  be  revolved  through  45°  so  that  its  axis  makes  an 
angle  of  45°  with  V,  in  the  same  direction  as  shown  in  Fig.  26, 
it  is  also  evident  that  the  surface,  which  makes  an  angle  of  40° 
with  H,  will  now  be  dark,  and  the  shade  lines  would  therefore 
change.  If  the  prism  be  revolved  through  45°  more  in  the 
same  direction,  its  axis  would  be  parallel  to  V,  and  the  surface 
in  question  would  then  be  light.  That  is,  the  surface  when 
perpendicular  to  V  would  be  light,  but  as  it  was  revolved  par- 
allel to  H,  at  some  intermediate  position  before  it  had  revolved 
45°,  it  would  become  dark,  changing  to  light  again  at  some 
intermediate  position  between  45°  and  90°  of  revolution. 

The  same  thing  would  occur  if  the  face  in  question  made  any 
angle  with  H  between  35°  16'  aud  45°.  The  foregoing  reason- 
ing would  apply  equally  well  to  the  under  face  «/'&/',  except  that 
this  one  would  be  dark  where  the  corresponding  upper  one 
would  be  light. 

Therefore,  if  a  surface  of  a  prism,  as  in  the  last  problem, 
mules  an  angle  with  H  between  35°  16'  and  45°,  that  surface 
becomes  doubtful  in  all  its  positions  lohen  the  axis  of  the  prism  is 
oblique  to  V,  and  the  shadow  of  this  surface  would  have  to  be 
cast  to  determine  positively  whether  it  is  light  or  dark. 

If  the  surfaces  ~make  angles  with  H,  not  included  between 
the  above  limits,  the  45°  triangle  on  the  end  view  would  deter- 
mine the  light  and  dark  surfaces  for  all  the  oblique  positions 
of  the  prism,  as  well  as  when  the  axis  is  perpendicular  to  V  or 
parallel  to  V  and  H. 

76.  Fig.  27  represents  the  two  elevations  and  plan  of  a  hol- 
low cylinder  whose  axis  is  parallel  to  V  and  H.     Here  the  end 


PROJECTIONS.  91 

elevation  would  naturally  be  drawn  first,  as  in  the  last  two  prob- 
lems, but  it  is  not  strictly  necessary,  as  both  of  its  projections, 
when  parallel  to  V  and  H,  are  the  same,  and  the  distance  apart 
of  the  contour  elements  is  equal  to  the  diameter  of  the  base. 

The  student  should  note  careful]}'  the  shade  lines  in  the  figure, 
especially  in  the  end  view. 

77.  Fig.  28  shows  the  plan  and  two  elevations  of  a  pile  of 
blocks.  The  lower  one  is  a  rectangular  prism,  the  second  one, 
which  rests  on  the  first,  is  the  frustum  of  a  square  pyramid,  and 
the  top  one  is  a  square  pyramid.  In  this  case  it  is  necessary 
to  draw  the  plan  first  and  construct  the  two  elevations  from  it, 
according  to  the  principles  already  explained. 

It  will  be  observed  that  the  group  is  considered  as  solid  in 
putting  in  shade  lines,  i.  e.,  the  edges  which  represent  the  peri- 
meter of  the  base  of  the  pyramid,  for  example,  are  considered 
as  separating  the  sloping  faces  of  the  pyramid  from  the  top 
surface  of  the  frustum  on  which  it  rests,  and  not  from  its  base, 
as  in  Fig.  23.  Compare  the  shade  lines  of  the  pyramids  in 
Figs.  23  and  28. 

Since  it  is  customary  to  tint  drawings  in  which  shadows  are 
cast,  shade  lines  would  not  be  put  in  on  the  same  drawing. 
Therefore,  it  is  advisable  to  disregard  the  shadows  altogether  in 
putting  in  shade  lines  on  line  drawings. 

78.  Prob.  13.  To  construct  the  projections  of  a  rigid  hex- 
agonal prism  when  its  axis  makes  a  compound  angle  with  the 
coordinate  plane,  the  angle  it  makes  with  H,  the  angle  the  hori- 
zontal projection  makes  tcith  GL,  and  the  size  of  the  prism  being 
given.    Figs.  29,  30,  and  31. 

There  are  three  distinct  steps  necessary  in  the  construction 
of  this  problem. 

Since  the  axis  is  oblique  to  both  planes  the  prism  cannot  be 


92  PROJECTIONS. 

drawn  in  the  required  position  directly,  but  must  be  placed  in 
such  a  position  as  will  show  two  dimensions.  Here  the  first 
step  is  to  draw  the  two  projections  of  the  prism  when  its  axis 
is  perpendicular  to  H  and  its  faces  make  the  required  angles 
with  V  (Art.  67).  Fig.  29  represents  the  projections  of  the 
prism  in  this  position. 

Next  draw  the  projections  of  the  prism  after  it  has  been 
revolved  in  the  proper  direction,  so  that  the  axis  makes  the 
correct  angle  with  H,  but  is  still  parallel  to  V.  Fig.  30  repre- 
sents the  projections  of  the  prism  in  this  position.  Since  the 
prism  keeps  a  constant  angle  with  V,  its  vertical  projection 
does  not  change  its  size  (Arts.  42-9th  and  54-4th).  Hence, 
make  the  vertical  projection  in  Fig.  30  the  same  size  as  in 
Fig.  29,  with  its  position  changed  so  that  the  elements  make 
the  correct  angle  with  GL.  In  this  revolution  no  point  changes 
its  distance  from  V.  Therefore,  to  construct  the  horizontal 
projection  of  the  prism  in  this  second  position,  draw  through 
each  corner,  as  av,  «_",  etc.,  Fig.  30,  lines  perpendicular  to  GL 
until  they  intersect  horizontal  lines  drawn  through  the  corre- 
sponding points,  as  ah  in  Fig.  29.  This  completes  the  second 
step. 

In  the  final  position  the  axis  is  to  make  the  same  angle  with 
H  as  it  does  in  the  position  just  drawn.  Hence  the  horizontal 
projection  must  be  the  same  size,  however  much  it  may  be 
revolved.  In  this  revolution  no  point  changes  its  height  above 
H.  Therefore,  draw  the  plan  of  the  prism,  Fig.  31,  the  same 
size  as  that  in  Fig.  30,  only  changing  the  angle  the  elements 
make  with  GL  the  required  amount.  Then  from  each  corner 
of  this  plan,  as  ah,ah,  etc.,  draw  perpendiculars  to  GL  until 
they  intersect  horizontal  lines  drawn  through  the  corresponding 
points,  av,a",  etc.,  in  Fig.  30.  Joining  these  corners  the  ver- 
tical projection  is  completed. 


Fig.13,  FjS-14 


Fig.  1 5 


Fig.  1 7 


Fig.  18 


r 1  1 

■r—  

^ *  : 


PROJECTIONS.  93 

If  the  angle  the  prism  made  with  V  and  the  angle  the  ver- 
tical projection  made  with  GL  had  been  given,  the  principles 
would  have  been  just  the  same,  only  you  would  have  first  drawn 
it  with  its  axis  perpendicular  to  V,  then  revolved  it  about 
a  vertical  axis  until  it  made  the  required  angle  with  V;  in 
this  case  the  plan  does  not  change  its  size  ;  and,  lastly,  revolved 
the  vertical  projection  last  found  through  the  proper  angle  and 
constructed  the  corresponding  plan. 

The  shade  lines  in  Figs.  30  and  31  can  only  be  determined 
positively  by  casting  the  shadows  of  the  doubtful  surfaces. 
Frequently  it  is  possible  to  tell  which  are  the  light  and  which 
the  dark  surfaces  without  casting  the  shadows  by  conceiving 
the  solid  in  its  position  in  space  together  with  the  ray  of  light, 
but  for  the  average  student  it  would  be  little  better  than  a  guess 
until  he  has  had  considerable  practice  in  finding  shadows. 

Before  taking  up  these  cases  where  it  is  necessary  to  cast  the 
shadow  in  order  to  determine  the  shade  line,  it  will  be  neces- 
sary to  take  up  so  much  of  the  subject  of  shadows  as  will  ena- 
ble the  student  to  find  the  shadow  of  an  ordinary  object  on  the 
two  coordinate  planes. 

79.  Piiob.  14.  To  construct  the  projections  of  a  right  hep- 
tagonal  pyramid  when  its  axis  makes  a  compound  angle  with  the 
coordinate  planes,  the  angle  it  makes  with  H,  the  angle  the  hori- 
zontal projection  makes  with  GL,  and  the  size  of  the  pyramid 
being  given. 

A  careful  examination  of  Figs.  32,  33,  and  34  will  be  suffi- 
cient to  understand  this  problem,  since  the  principles  are 
exactly  the  same  as  in  the  last  problem. 

80.  Prob.  15.  To  draw  the  projections  of  a  prism  1"  square 
and  If"  long,  resting  with  one  of  its  long  edges  on  H,  this  edge 


94  PROJECTIONS. 

making  an  angle  of  60°  with  V,  backward  and  to  the  right,  its 
front  end  being  2f "  in  front  of  V.  The  lower  left-hand  long 
face  making  an  angle  of  30°  with  H. 

Also,  draw  j)rojections  of  a  regular  hexagonal  pyramid  whose 
altitude  is  3",  diameter  of  circumscribed  circle  about  base  is  \\". 
The  lower  end  of  right-hand  element  of  pyramid  rests  on  II  1£" 
to  the  left  of  the  point  located  in  prism,  and  1-J-"  in  front  ofV; 
this  element  cdso  rests  on  top  edge  of  prism  at  a  point  £'  from  its 
front  end.  The  axis  of  pyramid  slopes  downward,  backtvard, 
and  to  the  left.  The  two  lower  faces  of  pyramid  make  equal 
angles  with  H.     Fig.  62. 

The  projections  of  the  prism  are  drawn  as  already  described 
in  Art.  74  ;  the  spaces  ahdh,  dhbh,  and  bhch  are  made  respectively 
equal  to  a  d,  db  ,  and  be. 

Heretofore,  in  drawing  the  projections  of  an  object  making 
a  compound  angle  with  the  coordinate  planes,  we  have  had  given 
the  size  of  the  object,  the  angle  it  made  with  V  or  H,  and  the 
angle  its  other  projection  made  with  GL,  and  the  object  has  been 
drawn  in  three  distinct  positions.  If  the  third  position  only  is 
wanted  it  is  not  essential  that  the  first  two  be  wholly  drawn, 
nor  that  they  be  made  in  separate  figures.  After  the  student 
becomes  familiar  with  this  work  so  that  numerous  lines  do  not 
confuse  him,  a  considerable  part  of  the  construction  may  be 
omitted.  In  the  figure  all  the  necessary  construction  lines 
have  been  left  in. 

Here  neither  the  angle  the  pyramid  makes  with  H,  nor  the 
angle  its  horizontal  projection  makes  with  GL,  are  given,  but 
the  projections  of  two  points  f  and  e  of  one  of  the  elements  are 
given,  which  enables  the  projections  of  this  element  to  be  drawn 
(indefinite  in  length).  Revolve  this  line  around  until  it  is  parallel 
to  V,  and  lay  off  on  it  in  this  position  the  true  length  of  the  ele- 
ment.    This  line  will,  of  course,  show  the  true  size  of  the  angle 


PROJECTIONS.  95 

this  element  makes  with  II,  and  the  horizontal  projection  of 
the  indefinite  line  hefore  revolution  shows  the  angle  it  makes 
with  GL.  All  the  necessary  data  are  now  obtained,  and  a  care- 
ful study  of  the  figure  should  enable  the  student  to  understand 
the  rest  of  the  construction. 

81.  It  is  evident  that  neither  the  height  of  an  object  above 
H  nor  its  distance  in  front  of  V  affects  in  the  least  the  size  and 
shape  of  its  projections.  Therefore,  the  GL  is  not  at  all  essen- 
tial in  drawing  the  projections  of  an  object  unless  its  distances 
from  V  and  II  are  given,  which  is  not  customary.  In  working 
drawings  the  GL  is  never  used.  It  is  only  used  in  elementary 
projections  as  an  aid  in  understanding  the  principles. 


CHAPTER  V. 


SHADOWS. 


82.  The  shadow  of  a  hody  upon  a  surface  is  that  portion 
of  the  surface  from  which  light  is  excluded  by  the  body. 

The  source  of  light  may  be  assumed  at  any  point,  but  it  is 
customary  to  assume  it  so  that  the  rays  of  light  are  parallel  to 
the  diagonal  of  a  cube,  as  already  stated  in  Art.  63,  in  which 
case  its  two  projections  make  angles  of  45°  with  GL.  Al- 
though rays  of  light  diverge  in  all  possible  directions  from  the 
source,  yet  when  this  source  is  as  far  removed  as  the  sun  there 
is  no  appreciable  error  in  calling  them  all  parallel. 

83.  The  shadow  of  a  point  on  any  surface  is  where  a  ray  of 
light  through  that  point  pierces  the  surface. 

Hence,  to  find  the  shadow  of  a  point  on  a  surface,  draw  a 
line  through  the  point  to  represent  the  ray  of  light,  and  find 
where  it  pierces  the  surface. 

84.  To  find  the  shadow  of  a  point  on  H. 

In  Fig.  35  let  b  represent  the  point  in  space  and  R  the  ray  of 
light  passing  through  this  point,  b"  and  b*1  will  be  the  two  pro- 
jections of  the  point  b  ;  bvr  and  Iflt  of  the  ray  of  light. 

The  shadow  of  the  point  ionH  will  be  where  R  pierces  H. 
This  point  being  in  H  will  have  its  vertical  projection  in  GL, 
and  its  horizontal  projection  will  be  the  point  itself  (Art.  33)  ; 
being  in  the  ray  of  light  its  two  projections  must  also  be  on 
the  projections  of  the  ray  of  light  (Art.  42-8th).  Therefore, 
produce,  if  necessary,  the  vertical  projection  of  the  ray  of  light 


SHADOWS.  97 

till  it  meets  GL  at  r,  and  r  will  be  the  vertical  projection  of  the 
point  where  the  ray  of  light  pierces  H ;  at  r  draw  a  perpendic- 
ular to  GL  and  %,  where  this  perpendicular  intersects  the  hori- 
zontal projection  of  the  ray  of  light  is  its  horizontal  projec- 
tion, and  is  the  shadow  of  the  point  b  on  the  plane  H. 
This  is  shown  in  actual  projection  in  Fig.  37. 

85.  To  find  the  shadow  of  a  point  on  V. 

In  Fig.  36  let  a  represent  the  point  and  R  the  ray  of  light 
passing  through  it.  a"  and  ah  will  be  the  projections  of  the 
point  a  ;  avr  and  aht  of  the  ray  of  light. 

The  shadow  of  the  point  a  on  V  will  be  where  R  pierces  V. 
This  point  being  on  V  will  have  its  horizontal  projection  in  GL, 
and  it  will  also  be  in  the  horizontal  projection  of  R,  hence  at 
their  intersection  t;  the  vertical  projection  of  this  point,  and 
the  shadow  on  V,  will  be  g£  where  a  perpendicular  drawn  from 
t  to  GL  intersects  a"r. 

This  is  shown  in  actual  projection  in  Fig.  38. 

86.  We  have  heretofore  supposed  that  the  coordinate  planes 
did  not  extend  below  or  behind  their  line  of  intersection,  but 
they  can  just  as  well  be  considered  as  extending  indefinitely  in 
both  directions,  as  shown  pictorially  in  Figs.  35  and  06.  Then, 
after  the  vertical  plane  has  been  revolved  to  coincide  with  the 
horizontal  plane,  that  portion  of  the  paper  above  GL  repre- 
sents not  only  that  portion  of  V  which  is  above  H,  but  that  part 
of  H  which  is  behind  V ;  also  that  portion  of  the  paper  below 
GL  represents  that  part  of  H  which  is  in  front  of  V  and  that 
part  of  V  which  is  below  H. 

Referring  again  to  Fig.  35,  we  have  already  seen  that  R 
pierces  H  at  k*;  now,  if  we  suppose  R  to  be  produced  below  H 
indefinitely,  it  must  pierce  V  at  some  point,  since  it  is  not  par- 
allel to  V.  This  point  is  found  in  exactly  the  same  way  as 
already  described  in  Art.  85.     It  does  not  make  a  particle  of 


98  SHADOWS. 

difference  whether  it  pierces  V  above  or  below  H.  That  is, 
every  ray  of  light,  unless  parallel  to  V,  will  pierce  V  at  some 
point  either  above  or  below  H,  and  since  these  points  are  all 
in  V  their  horizontal  projections  must  be  in  GL.  Of  course, 
the  shadow  of  the  point  b  falls  on  H,  and  does  not  actually  fall 
on  V,  but  the  point  can  be  found  where  it  would  fall  if  H  were 
transparent,  and  it  is  frequently  convenient  to  do  this  in  finding 
shadows  of  bodies  in  certain  positions,  as  we  shall  soon  see. 

Fig.  37  shows  this  point  b,  in  actual  projection.  It  being 
on  that  part  of  V  which  is  below  H  after  revolution  appears 
below  GL. 

Referring  again  to  Fig.  36,  it  is  evident  that  R  not  only 
pierces  V  at  a",  but  also  pierces  H  at  a^.  ahs  being  in  H  has 
its  vertical  projection  in  GL. 

Fig.  38  shows  this  point  in  actual  projection.  It  being  on 
that  part  of  H  which  is  behind  V,  after  revolution  appears 
above  GL. 

87.     The  following  rules  are  evident  from  the  forea-oina: :  — 

To  find  the  shadow  of  a  point  on  H,  produce  the  vertical  pro- 
jection of  the  ray  of  light  to  meet  GL ;  erect  a  perpendicular  at 
this  point  of  intersection,  and  the  intersection  of  this  perpen- 
dicular with  the  horizontal  projection  of  the  ray  of  liyht  will  be 
the  shadow  required. 

It  should  be  carefully  borne  in  mind  that  this  last  perpendic- 
ular may  intersect  the  horizontal  projection  of  the  ray  of  light 
above  or  below  GL,  depending  on  the  location  of  the  point  in 
space ;  that  is,  if  the  point  be  nearer  H  than  V  the  intersection 
will  be  below  GL,  and  the  shadow  will  actually  fall  on  H ;  if 
the  point  be  nearer  V  than  H  the  intersection  will  be  above  GL, 
and  the  shadow  will  be  imaginary. 

To  find  the  shadow  of  a  point  on  V,  produce  the  horizontal 
projection  of  the  ray  of  light  to  meet  GL ;  erect  a  perpendicular 


SHADOWS.  99 

at  this  point  of  intersection,  and  the  intersection  of  this  perpen- 
dicular with  the  vertical  projection  of  the  ray  of  light  will  be  the 
shadow  required. 

Here,  also,  the  same  caution  as  for  the  last  rule  is  applicable. 
But  in  this  case  if  the  point  is  nearer  V  than  H  the  intersec- 
tion is  above  GL,  and  the  shadow  actually  falls  on  V ;  if  the 
point  is  nearer  H  than  V  the  intersection  will  be  below  GL, 
and  the  shadow  will  be  imaginary. 

The  following  may  also  be  noted  :  — 

If  the  horizontal  projection  of  the  ray  of  light  meets  the  GL 
before  the  vertical  projection,  the  shadow  will  actually  fall  on  V; 
if  the  vertical  projection  meets  the  GL  before  the  horizontal,  the 
shadow  actually  falls  on  H. 

88.  Fig.  39  shows  how  to  find  the  shadow  of  a  line  par- 
allel to  both  V  and  H. 

Fig.  40  shows  the  shadow  of  a  line  perpendicular  to  and 
resting  on  H. 

Fig.  41  shows  the  shadow  of  a  line  perpendicular  to  H  but 
not  resting  on  H.  In  this  case  a  part  of  the  shadow  falls  on 
each  of  the  coordinate  planes. 

From  these  figures  the  following  facts  may  be  noted :  — 

1st.  The  shadow  of  a  straight  line  on  a  plane  surface  is  a 
straight  line. 

2nd.  The  shadow  of  a  line  on  a  plane  to  which  it  is  paral- 
lel is  a  line  parallel  and  equal  to  it  in  length. 

3rd.  The  shadow  of  a  line  on  a  plane  to  which  it  is  perpen- 
dicular coincides  with  the  projection  of  the  ray  of  light  on  that 
plane,  and  it  is  longer  than  the  line  itself. 

4th.  7  he  shadow  of  a  line  on  a  plane  may  be  said  to  begin 
where  the  line  pierces  that  plane,  either  or  both  being  produced 
if  necessary. 

5th.     Since  two  points  determine  a  straight  line  it  is  sufficient 


100  SHADOWS. 

to  find  the  shadow  of  two  points  of  it  on  a  plane  surface.  In 
case  the  direction  of  the  shadow,  or  the  point  where  the  line  meets 
the  plane  surface  receiving  the  shadow  is  known,  it  is  sufficient 
to  construct  the  shadow  of  one  other  point  only. 

Gth.  When  the  shadow  of  a  line  falls  upon  two  surfaces  winch 
intersect,  the  shadows  on  the  ttco  surfaces  meet  at  a  common  point 
on  their  line  of  intersection.  This  is  equally  true  whether  the 
two  surfaces  intersect  at  right  angles  to  each  other  or  otherwise. 

89.  Prob.  15.  To  find  the  shadow  in  V  and  H  of  a  line 
oblique  to  V  and  H,  when  one  end  is  nearer  H  than  V,  and  the 
other  is  nearer  V  than  H.     Fig.  42. 

Let  arlf  and  ahVl  be  the  two  projections  of  such  a  line.  The 
end  a  being  nearer  H  than.V  its  shadow  will  fall  on  H  (Art. 
87),  and  will  be  found,  as  already  described,  at  a*.  The  end  b, 
being  nearer  V  than  H,  its  shadow  will  fall  on  V,  and  will  be 
found  at  6".  Since  the  shadow  of  one  end  falls  on  H  and  of 
the  other  on  V,  it  is  evident  that  the  shadow  of  the  line  will 
fall  partly  on  H  and  partly  on  V,  and  also  that  a  line  joining 
these  two  points  could  only  be  a  line  in  space,  and  therefore 
not  the  shadow  required. 

It  is  essentia]  that  the  two  points  which  determine  the  shadow 
of  a  line  should  be  on  one  and  the  same  plane  ;  therefore,  as  we 
have  the  shadow  of  one  point  of  the  line  on  each  coordinate 
plane,  it  is  necessary  to  construct  the  shadow  of  another  point 
of  the  line  on  either  of  the  coordinate  planes.  Any  point  could 
be  taken,  but  the  ends  being  definitely  projected  it  is  more  con- 
venient to  use  them.  We  have  already  found  the  shadow  of  a  on 
H  to  be  a* ;  the  shadow  of  b  on  H  is  found  at  #'',  in  the  same 
way.  (This  shadow  we  know  does  not  actually  fall  on  H,  but 
it  serves  our  purpose,  which  is  to  get  the  direction  of  the  shadow 
of  the  line,  just  as  well  as  if  it  did)  ;    ahsbbs  is,  therefore,  the 


SHADOWS.  101 

shadow  of  the  whole  line  on  H,  hut  only  that  part  of  it,  ahsc, 
which  falls  below  or  in  front  of  GL  is  actual  shadow.  "We 
have  also  seen  that  the  shadow  of  b  on  V  is  #,',  the  shadow  of  a 
on  V  is  similarly  found  at  oj,  and  a'^  is  the  shadow  of  the 
whole  line  on  V,  but  only  that  part  of  it,  hvsc,  which  falls  above 
or  behind  GL  is  actual  shadow.  We,  therefore,  have  as  the 
actual  shadow  of  the  line  a''c  on  H  and  b'sc  on  V,  the  portions 
on  the  two  planes  intersecting  in  a  common  point,  c,  on  GL  as 
already  noted  in  Art.  88-6th.  Since  this  is  the  case  it  is  evi- 
dent that  it  is  not  necessary  to  find  the  shadow  of  the  whole 
line  on  both  V  and  H ;  having  found  the  point  c  where  either 
shadow  crosses  GL,  join  this  point  with  the  shadow  of  the  end 
which  falls  on  the  other  plane. 

90.  Fig.  43  shows  the  shadow  on  V  of  a  square  card  whose 
surface  is  parallel  to  V.  Since  the  edges  of  the  card  are  parallel 
to  V  their  shadows  will  he  parallel  and  equal  each  to  each,  and 
consequently  the  shadow  of  the  card  will  be  equal  and  parallel 
to  the  card  itself.  This  will  be  true  whatever  the  size  or  shape 
of  the  card. 

Therefore,  the  shadow  of  any  plane  figure  on  a  surface  to 
which  it  is  parallel  is  a  figure  equal  and  parallel  to  it. 

91.  Fig.  44  shows  the  shadow  on  V  of  a  square  card  whose 
surface  is  perpendicular  to  V  and  parallel  to  H.  Here  the  two 
edges  ah  and  cd  are  parallel  to  V,  consequently  their  shadows 
will  be  equal  and  parallel  lines.  The  other  two  edges,  ad  and 
&c,  are  perpendicular  to  V,  hence  their  shadows  make  angles  of 
45°  with  GL. 

92.  Fig.  45  shows  the  shadow  on  both  V  and  H  of  a  square 
card  whose  surface  is  perpendicular  to  both  V  and  H,  that  is, 
is  in  a  profile  plane.  The  edge  cd  is  parallel  to  V,  and  is  nearer 
to  V  than  H,  hence  its  shadow  is  on  V  parallel  and  equal  to 
cd;  the  edge  ad  is  nearer  to  V  than  H,  and  is  perpendicular  to 


102  SHADOWS. 

V,  hence  its  shadow  is  on  V,  and  makes  an  angle  of  45°  with 
GL  ;  the  edge  ab  is  parallel  to  V  and  perpendicular  to  H,  the 
upper  end  is  nearer  V  than  H,  and  the  lower  end  is  nearer  H 
than  V,  hence  its  shadow  is  partly  on  V  and  partly  on  H ;  that 
portion  which  falls  on  V  will  be  parallel  to  arb",  and  that  por- 
tion which  falls  on  H  will  make  an  angle  of  45°  with  GL ;  the 
other  edge  be  is  parallel  to  H  and  perpendicular  to  V,  the  front 
end  is  nearer  H  than  V,  and  the  back  end  is  nearer  V  than  H, 
hence  its  shadow  will  fall  partly  on  V  and  partly  on  H ;  that 
portion  which  is  on  H  will  be  parallel  to  b''ch,  and  that  portion 
which  is  on  V  will  make  an  angle  of  45°  with  GL.  The 
shadow  of  the  whole  card  on  V  is  a^c^d^,  of  which  only  that 
portion  a'snmc'sd's,  which  is  above  GL,  is  visible.  The  shadow 
of  the  whole  card  on  H  is  a*b%c*ds,  of  which  only  that  portion 
b*mn  is  visible.  It  is,  of  course,  not  essential  to  find  that  por- 
tion ahsnmc*d's  of  the  shadow  on  H  which  is  above  GL. 

93.  Fig.  46  shows  the  shadow  of  a  card  lying  in  a  profile 
plane,  all  of  its  edges  being  oblique  to  both  V  and  H.  Each 
point  being  found  the  same  as  all  the  preceding  ones,  no  further 
explanation  is  necessary. 

94.  Fig.  47  shows  the  shadow  of  a  circular  card  parallel  to 
H.  The  shadow  on  H  we  know  must  be  a  circle  equal  in  size 
to  the  card,  therefore  it  is  only  necessary  to  find  the  shadow  of 
its  centre  o.  This  is  found  at  o''.  With  this  point  as  a  centre 
and  a  radius  equal  to  that  of  the  card  describe  the  arc  of  a 
circle  mm.  Since  a  part  of  the  circle  is  above  GL  it  is  evident 
that  that  part  of  the  shadow  actually  falls  on  V.  To  get  this 
shadow  take  any  points  on  the  circle,  as  a,b,c,d,  etc.,  and  fiud 
their  shadows  separately,  joining  these  points  by  a  curved  line. 
The  points  m  and  n,  where  the  circle  described  from  o*  as  a  centre 
crosses  GL,  will,  of  course,  be  two  points  on  the  curve.  This 
curve  will  be  an  ellipse,  and  the  shadoio  of  a  circle  on  a  plane 
to  which  it  is  perpendicular  or  oblique  will  be  an  ellipse. 


SHADOWS.  103 

95.  As  a  solid  is  composed  of  planes,  planes  of  lines,  and 
lines  of  points,  it  is  evident  that  the  shadow  of  the  most  com- 
plex body  is  obtained  by  finding  the  shadow  of  one  point  at  a 
time  by  the  methods  already  given  until  the  shadows  of  all  the 
points  on  the  object  which  cast  shadows  have  been  found,  so 
that  the  student  who  finds  himself  now  able  to  cast  the  shadow 
of  any  single  point  on  a  given  plane  has  practically  mastered 
the  subject,  and  if  such  a  one  has  any  difficulty  in  finding  the 
shadow  of  any  object  the  trouble  is  that  he  does  not  understand 
thoroughly  the  subject  of  projections. 

96.  Since  the  shade  lines  of  a  body  separate  its  light  from 
its  dark  surfaces,  the  shadow  of  the  shade  lines  will  form  the 
boundary  of  the  shadow  of  the  body.  Therefore,  in  finding 
the  shadow  of  a  body  the  shade  lines  should  first  be  marked, 
if  it  is  in  such  a  position  that  they  can  be  found  by  means  of 
the  45°  triangle,  and  the  shadows  of  these  lines  give  the  shadow 
of  the  whole  object.  If  the  object  is  in  such  a  position  that 
the  shade  lines  cannot  be  found  by  means  of  the  45°  triangle 
directly,  the  shadow  of  every  point  on  the  object,  except  those 
which  it  is  known  do  not  cast  shadows,  should  be  found  sepa- 
rately, and  then  join  those  points  which  will  enclose  the  largest 
area.  The  shade  lines  can  then  be  found  from  the  boundary 
of  the  shadow  by  finding  what  lines  on  the  object  cast  these 
boundary  lines. 

97.  Prob.  16.  To  find  the  shadow  of  an  hexagonal  prism 
with  its  two  ends  parallel  to  II.     Fig.  48. 

The  shade  lines  of  the  prism  in  this  position  are  found 
directly  by  means  of  the  45°  triangle  to  be  ab,  be,  cd,  de,  ef, 
fm,  mn,  and  an.  It  is  only  necessary  to  find  the  shadow  of 
each  of  these  lines  and  join  them  in  order,  and  the  shadow  is 
completed.     The  first  three  and  last  three  of  these  lines  are 


104  SHADOWS. 

parallel  to  H,  hence  their  shadows  on  II  will  be  equal  and  par- 
allel respectively  to  the  edges  casting  them. 

98.  Fig.  49  shows  the  shadow  of  a  square  prism  on  V  and 
H,  resting  with  its  base  on  H  and  its  long  faces  oblique  to  V. 
The  shade  lines  are  found  first  here. 

99.  Fig.  50  shows  the  shadow  of  a  cylinder  on  V  and  H, 
with  its  base  resting  on  H.  The  shadow  of  any  number  of 
points  on  the  shade  line  between  a  and  d  can  be  found.  That 
portion  of  the  curve  between  a's  and  dl  will  be  a  semi-ellipse, 
and  the  lines  avsm  and  d'sn  must  be  tangent  to  this  ellipse  at  the 
points  a's  and  d%. 

100.  Pkob.  17.  To  find  the  shadow  of  a  right  cone  resting 
with  its  base  on  H.     Fig.  51. 

It  is  evident  that,  unless  all  the  sloping  part  of  the  cone  is 
light,  there  will  be  two  elements  of  the  cone  which  separate 
light  from  dark  surfaces,  and  also  that  these  two  elements  meet 
at  the  vertex  and  terminate  at  the  other  end  in  the  base.  But 
we  do  not  know  just  where  the  light  surface  stops  and  the  dark 
surface  begins,  as  we  do  in  the  case  of  the  cylinder,  so  we  have 
to  cast  the  shadow  first. 

The  shadow  of  the  vertex  o  is  oj ;  from  o]  draw  two  lines 
ohsah  and  ohsbh  tangent  to  the  base  of  the  cone,  and  these  will  be 
the  shadows  of  the  two  shade  elements.  We  now  see  that  oa 
and  ob  are  the  dividing  lines  between  the  light  and  dark  sur- 
faces, but  since  these  lines  do  not  coincide  with  the  contour 
elements  od  and  oc,  it  is  evident  that  neither  od  nor  oc  is  a  shade 
line. 

The  bottom  of  the  cone  is  dark  and  the  sloping  surface  ohahchbh 
is  light,  hence  the  edge  a^cW  is  a  shade  line. 

Note  the  difference  in  the  shade  lines  on  the  cylinder  and 
the  cone. 


SHADOWS.  105 

Do  not  shade  the  contour  elements  of  a  cylinder  or  cone. 

101.  We  have  seen  that  to  find  the  shadow  of  a  body  three 
things  must  be  given,  the  body  casting  the  shadow,  the  surface 
receiving  the  shadow,  and  the  ray  of  light,  and  that  these  must 
be  given  by  their  projections.  The  surface  receiving  the  shadow 
has  for  convenience  so  far  been  taken  as  one  of  the  coordinate 
planes,  one  projection  of  which  is  the  plane  itself  and  the  other 
is  the  ground  line,  but  it  is  more  frequently  necessary  to  find 
shadows  of  objects  upon  themselves  and  upon  other  objects  in 
the  immediate  vicinity.  We  have  also  seen  that  in  finding 
shadows  on  the  coordinate  planes  we  have  only  concerned  our- 
selves with  that  projection  of  the  surface  which  is  a  line,  that 
is,  the  GL.  Hence,  to  find  the  shadow  of  an  object  on  surfaces 
other  than  the  coordinate  planes,  we  have  only  to  find  the  line 
which  corresponds  with  GL  and  proceed  according  to  the  rule 
already  given  (Art.  87). 

To  find  this  line,  which  we  may  call  GL,  observe  the  follow- 
ing rule  :  — 

Rcle  for  GL.  The  GL  to  be  used  in  finding  shadows  is 
always  that  projection,  of  the  surface  receiving  the  shadow  which 
is  a  line.      This  line  may  be  straight,  curved,  or  otherwise. 

Of  course,  this  rule  is  applicable  only  when  one  projection  of 
the  surface  is  a  line. 

102.  Prob.  18.  To  find  the  shadow  of  a  slick  of  timber  on 
the  top  and  front  of  an  abutment  on  which  it  rests.     Fig.  52. 

The  shade  lines  of  the  stick  of  timber  are  easily  found  to  be 
as  follows :  cdt,  dd,  de,  and  ef  First  find  the  shadow  on  the 
top  surface  of  the  abutment ;  the  vertical  projection  of  this  sur- 
face is  the  line  arbv,  which,  according  to  the  rule,  is  the  GL  to 
be  used.  The  shadow  of  dd  is  dhd\,  of  de  is  d^e'l,  and  of  ef 
(since  e/*is  parallel  to  the  top  of  the  abutment)  is  a  line  through 


106  SHADOWS. 

e*  parallel  to  ehfh.  Next  find  the  shadow  on  the  front  surface 
of  the  abutment ;  the  horizontal  projection  of  this  surface  is 
the  line  ahbh,  which  must  therefore  be  the  GL  for  this  surface. 

The  shadow  of  the  shade  edge  ef  will  fall  on  both  the  top 
and  front  surfaces,  which  intersect  in  the  line  ab,  therefore  the 
shadows  on  the  two  surfaces  will  meet  in  a  common  point  m  on 
this  line  (Art.  88-6th).  mv  is,  then,  one  point  of  the  shadow 
of  the  edge  ef  on  the  front  of  the  abutment.  The  point  np,  for 
the  same  reason,  is  one  point  of  the  shadow  of  the  edge  cd  on 
this  surface.  Since  these  two  edges  are  parallel  they  will  cast 
parallel  shadows,  hence  it  is  only  necessary  to  find  the  shadow 
of  one  more  point,  and  the  shadow  is  determined.  The  shadow 
of  the  point  c  is  c£ ;  join  this  with  nv,  and  through  mv  draw  a 
line  parallel  to  c"n",  and  the  shadow  is  completed.  Of  course, 
any  other  point  might  have  been  taken  on  the  edge  cdt,  or  any 
point  on  the  edge  ef 

The  irregular  line  r!ish  simply  indicates  that  the  abutment 
extends  backward  farther  than  it  was  necessary  to  show.  The 
ragged  end  of  the  stick  of  timber  indicates  the  same  thing. 

103.  Prob.  19.  To  find  the  shadow  of  a  stick  of  timber 
on  another  stick  of  timber  into  which  it  is  framed.     Fig.  53. 

It  is  evident  that  the  sloping  stick  will  cast  a  shadow  on  the 
top  of  the  horizontal  stick.  The  lower  front  and  the  upper  back 
edges  cd  and  ab  are  the  lines  which  cast  the  shadows,  and  mn  is 
the  line  to  be  used  as  GL.  The  shadow  of  the  point  a  is  at 
a*,  apparently  on  the  elevation  of  the  object,  but  in  reality  it  is 
where  the  shadow  would  be  if  the  lower  stick  of  timber  were 
sufficiently  wide  to  receive  it.  It  is  frequently  necessary  to 
imagine  surfaces  indefinite  in  extent  for  convenience  in  con- 
struction. The  shadow  of  the  edge  ab  begins  where  it  leaves 
the  surface   (Art.  88-4th),  therefore,  join  ah  and  bh  and  that 


SHADOWS.  107 

portion  of  the  line  which  falls  on  the  surface  of  the  stick  will 
be  all  that  is  necessary  ;  through  ch  draw  a  line  parallel  to  V'ah, 
and  the  shadow  is  completed. 

104.  Prob.  20.  To  find  the  shadoio  of  one  oblique  stick  of 
timber  on  another,  both  being  parallel  to  V  and  lying  against  each 
other.     Fig.  54. 

The  shadow  will  fall  on  the  front  and  top  of  the  back  stick. 
The  ground  line  for  finding  the  shadow  on  the  front  surface  is 
rs,  and  for  the  top  surface  is  mn.  Here,  as  in  Prob.  18,  the 
horizontal  projections  of  the  points  cv  aud  dv,  where  the  shadow 
leaves  the  front  surface,  will  be  points  of  the  shadow  on  the  top. 
A  careful  examination  of  the  figure  will  make  further  explana- 
tion unnecessary. 

105.  Prob.  21.  To  find  the  shadow  of  a  straight  wire 
lying  on  top  of  a  vertical  cylindrical  wall  on  the  wall  and  the 
shadow  of  the  wall  on  itself     Fig.  55. 

Here  the  GL  is  a  curved  line,  and  is  to  be  used  just  the  same 
as  heretofore.  There  being  no  new  principles  the  student  should 
be  able  to  understand  this  problem  from  an  examination  of  the 
figure. 

106.  Prob.  22.  To  find  the  shadoto  of  the  head  of  a  bolt 
on  its  shank  when  the  length  of  the  bolt  is  parallel  to  both  V 
and  H. 

In  Fig.  56  the  head  is  hexagonal  and  the  shank  is  cylindri- 
cal, and  in  Fig.  57  the  head  is  octagonal  and  the  shank  is 
hexagonal. 

In  cases  like  these  it  is  evident  that  neither  the  plan  nor  ele- 
vation of  the  surfaces  receiving  the  shadow  is  a  line,  but  the  end 
view,  or  the  projection  on  a  profile  plane,  is  a  line,  and  accord- 


108  SHADOWS. 

ing  to  the  rule  is,  therefore,  the  GL  to  be  used,  mno  is  the  GL 
in  Fig.  56,  and  mnor  in  Fig.  57.  The  only  other  new  point  to 
be  noticed  is  that  the  two  elevations  of  the  ray  of  light  each 
make  an  angle  of  45°  with  a  horizontal  line,  and  slope  in  the 
same  direction. 

107.  Prob.  23.  To  find  the  shadow  of  a  chimney  or  tower 
of  a  house  on  the  roof     Fig.  58. 

The  end  view  of  the  roof  mn  is  the  GL  for  this  problem. 
The  shadow  is  found  the  same  as  in  Figs.  56  and  57. 

108.  Prob.  24.  To  find  the  shadow  of  a  stick  of  timber  on 
the  top  and  sloping  faces  of  an  oblique  abutment.     Fig.  59. 

Since  neither  plan,  elevation,  nor  end  view  of  these  sloping 
surfaces  is  a  line  our  rule  is  no  longer  directly  applicable,  and 
we  may  make  use  of  an  indirect  method  (the  shadow  can  be 
found  directly  by  descriptive  geometry  methods). 

First  find  the  shadow  on  top  of  the  abutment.  The  points 
ch  and  eh,  where  it  leaves  the  top,  will  be  two  points  of  the 
shadow  on  the  left,  front,  sloping  face  (Art.  88-6th),  cv  and  ev 
will  be  their  vertical  projections ;  in  the  same  way  dh  and  nh 
will  be  two  points  of  the  shadow  on  the  right,  front,  sloping  face, 
and  dv  and  nv  will  be  their  vertical  projections.  The  shadow 
of  the  lower  front  edge  (the  upper  back  edge  could  just  as  well 
have  been  taken)  on  the  ground  is  the  indefinite  line  fhoh.  The 
ground  and  the  sloping  face  of  the  abutment  intersect,  hence 
the  pointy*  will  be  a  point  of  the  shadow  on  the  left  front  face  ; 
f"  will  be  its  vertical  projection  ;  for  the  same  reason  oh  is  a 
point  of  the  shadow  on  the  right,  front  face,  and  ov  is  its  verti- 
cal projection.  Join  e'fh,  and  through  ch  draw  a  line  parallel 
to  it ;  also,  nhoh  and  a  line  through  dh  parallel  to  it.  This  com- 
pletes the  shadow  on  the  plan.  The  shadow  on  the  elevation 
is  found  by  projection  from  the  plan. 


SHADOWS.  109 

Ii  the  abutment  had  been  so  high  that  the  shadow  on  the 
ground  would  have  been  difficult  to  obtain,  an  imaginary  hori- 
zontal plane  could  have  been  taken  at  any  convenient  place, 
and  the  shadow  found  on  that,  noting  where  it  comes  out  from 
the  abutment.  The  dash  and  two  dots  line  in  plan  and  eleva- 
tion represents  the  two  projections  of  the  intersection  of  such 
an  imaginary  plane  with  the  abutment.  m,'t'1  is  the  shadow  of 
the  lower  front  edge  of  the  stick  on  this  imaginary  plane,  which 
gives  the  points  tf1  and  mh  as  points  of  the  shadow.  The  shadow 
is  completed  as  before. 

109.  Prob.  25.  To  find  the  shadow  of  any  oblique  line  on 
any  oblique  plane.     Fig.  60. 

First  find  the  shadow  on  any  horizontal  auxiliary  plane. 
mvev  will  be  the  vertical  projection  of  one  such  plane,  and 
will  be  the  GL  for  this  plane.  The  shadow  of  any  point  as  a 
on  this  plaue  will  be  ahs ;  the  shadow  of  any  other  point  could 
be  found  in  the  same  way,  thus  getting  the  shadow  of  the  whole 
line,  but  the  shadow  of  a  line  begins  on  a  plane  where  the  line 
pierces  the  plane,  and  the  line  pierces  this  auxiliary  plane  at 
the  point  b  (b°  being  where  mrev  intersects  the  vertical  projec- 
tion of  the  line,  and  I/1  being  perpendicularly  below  it  on  the 
horizontal  projection  of  the  line),  therefore,  joining  a,  and  V1 
we  have  the  shadow  on  the  auxiliary  plane.  mvev  is  the  vertical 
projection  of  the  line  of  intersection  of  the  auxiliary  plane  with 
the  card  mnop  ;  its  horizontal  projection  is,  therefore,  mheh.  The 
lines  Z>V''  and  mheh  lie  in  the  auxiliary  plane,  the  line  mheh  also 
lies  in  the  plane  of  the  card  mnop,  therefore,  the  point  rh,  where 
these  two  lines  intersect,  is  a  point  on  the  card,  and  must  be  one 
point  of  the  horizontal  projection  of  the  shadow  required  ;  rv  on 
mvev  is  its  vertical  projection.  Assume  any  other  auxiliary 
pla.ie,   as  dvov,  and  another  point,  svsh  of  the  shadow,  will  be 


110  SHADOWS. 

found  in  the  same  way.     Draw  an  indefinite  line  through  these 
points  r  and  s  in  both  projections,  and  the  shadow  is  finished. 

Vertical  auxiliary  planes  could  have  been  taken  instead  of 
horizontal  with  the  same  result. 

110.  Prob.  26.  To  cast  the  shadoio  of  an  abacus  on  a  con- 
ical column.     Fig.  61. 

Since  neither  projection  of  the  surface  receiving  the  shadow 
is  a  line,  this  problem  must  be  done  by  the  indirect  method  as 
used  in  the  two  preceding  problems. 

Find  the  shadow  on  any  horizontal  plane,  as  avcv.  Its  ver- 
tical projection  is  avcv,  and  is  the  GL  for  this  shadow.  This 
plane  cuts  the  column  in  a  horizontal  circle,  of  which  avcv  is 
the  vertical  and  aJ'V'c'1  is  the  horizontal  projection.  The  shadow 
of  the  bottom  edge  of  the  abacus  on  this  plane  is  the  circle 
dhaheh,  drawn  with  o*,  the  shadow  of  o  the  centre  of  the  abacus, 
as  a  centre  and  radius  equal  to  that  of  the  abacus.  This  circle 
cuts  the  circle  aft5*c*  at  the  point  ah,  which  is  one  point  of  the 
horizontal  projection  of  the  shadow  required.  av,  on  the  verti- 
cal projection  of  this  auxiliary  horizontal  plane,  is  one  point  of 
the  vertical  projection  of  the  required  shadow.  Any  number 
of  other  points  can  be  found  in  the  same  way. 

111.  Prob.  27.  To  cast  the  shadow  of  the  prism  given  as 
in  Fig.  62  on  H  and  V,  also  of  the  pyramid  on  the  prism  and 
on  H. 

The  shadows  of  the  prism  on  H  and  V  and  of  the  pyramid 
on  H  require  no  explanation. 

If  the  shadow  of  a  line  falls  partly  on  two  plane  surfaces, 
A  and  B  (no  figure),  which  do  not  intersect  (at  that  portion 
under  discussion),  A  being  between  the  line  and  B,  that  part 
which  falls  on  A  does  not  fall  on  B,  and  the  shadow  of  the  line 


SHADOWS.  Ill 

on  B  may  be  said  to  begin  at  the  shadow  of  the  point  where 
the  shadow  of  the  line  leaves  A,  on  B.  Now,  if  B  is  horizontal 
or  vertical,  and  A  is  oblique  to  both  V  and  H,  the  shadows  of 
the  line  and  the  plane  A  on  B  can  be  readily  found ;  and  then 
get  the  shadow  of  the  line  on  A  by  the  reverse  of  the  above 
process,  that  is,  note  where  the  shadow  of  the  line  and  plane  A 
on  B  intersect ;  find  what  point  on  A  cast  this  intersecting 
point,  and  that  will  be  one  point  of  the  shadow  of  the  line  on  A. 

Now,  referring  to  Fig.  62,  we  see  that  the  shadow  of  the 
pyramid  and  prism  on  H  intersect  at  the  four  points  rj,  ^,  x%, 
and  zhs  ;  the  points  on  the  prism  (this  being  between  the  pyra- 
mid and  H)  which  cast  these  shadows  are  rh,  th,  xh,  and  zh 
respectively  (found  by  drawing  a  45°  line  from  the  shadow  to 
the  edge  casting  it),  and  they  are,  therefore,  points  of  the 
shadow  of  the  pyramid  on  the  prism. 

The  shadow  on  the  upper  edge  of  the  prism  may  be  found 
by  this  same  principle,  that  is,  the  shadow  of  this  upper  edge 
on  H  is  bltf* ;  this  intersects  the  shadow  of  the  pyramid  on  H 
at  s'g  and  yhs  ;  sh  and  yh  are  the  points  which  cast  these  shad- 
ows, hence  they  are  points  of  the  shadow  required.  Join  rhsh, 
shth,  xhyh,  and  yhzh,  and  the  horizontal  projection  of  the  shadow 
is  completed.  Its  vertical  projection  is  found  by  projecting 
each  one  of  these  poiuts  on  to  the  vertical  projection  of  the 
prism.  The  points  sh  and  yh  could  also  have  been  found  by 
casting  the  shadow  of  the  pyramid  on  an  auxiliary  horizontal 
plane  through  the  top  edge  of  the  prism ;  o£sh  and  o*yh  are  the 
shadows  of  the  two  shade  elements  of  the  pyramid  on  such  a 
plane ;  they  intersect  the  upper  edge  of  the  prism  at  the  points 
sh  and  y'\  which  are  the  points  required.  A  vertical  auxiliary 
plane  through  this  edge  or  the  other  edges  of  the  prism  might 
have  been  used  with  the  same  result. 


CHAPTER  VL 


ISOMETRICAL    DRAWING. 


112.  In  all  the  previous  constructions  two  projection.?  have 
been  used  to  represent  a  body  in  space.  In  isometrical  projec- 
tions only  one  view  is  used,  the  body  being  placed  in  such  a 
position  that  its  principal  lines  or  edges  (length,  breadth,  and 
thickness)  are  parallel  to  three  rectangular  axes,  which  are  so 
placed  that  equal  lengths  on  them  are  projected  on  the  plane 
equal  to  each  other.  Thus  we  have  the  three  dimensions  of  a 
body  shown  on  one  plane  in  such  a  way  that  each  can  be  meas- 
ured, thereby  combining  the  exactness  of  ordinary  projections 
and  the  intelligibleness  of  pictorial  figures.  It  is  used  chiefly 
to  represent  small  objects  in  which  the  principal  lines  are  at 
right  angles  to  each  other.  In  large  objects  the  drawing  would 
look  distorted. 

113.  If  we  take  a  cube  situated  as  in  Fig.  63,  and  tip  it  to 
the  left,  about  its  lower  left  corner  e  until  the  diagonal  eg  is 
horizontal,  Fig.  64,  and  then  turn  it  through  an  angle  of  90°, 
still  keeping  eg  horizontal,  we  obtain  Fig.  65.  The  vertical 
projection  in  this  figure  is  what  is  called  an  isometrical  projec- 
tion. 

The  edges  of  a  cube  are  all  of  equal  length,  and  it  will  be 
seen  that  they  appear  equal  in  the  figure,  consequently  the  visi- 
ble faces  must  appear  equal.  It  will  also  be  seen  that  the  figure 
can  be  inscribed  in  a  circle,  and  that  the  outline  of  the  isomet- 
rical  projection  i»  a   regular  hexagon,  hence,  that  those  lines 


V 


c 


ISOMETRICAL  DRAWING.  113 

which  represent  length  and  breadth   make  angles  of  30°  with 
a  horizontal,  and  those  which  represent  thickness  are  vertical. 

114.  The  edges  of  the  cube  being  inclined  to  the  plane  on 
which  they  are  represented,  appear  shorter  than  they  actually 
are  on  the  object,  but  since  they  are  all  equally  foreshortened, 
and  since  a  drawing  may  be  made  to  any  scale,  it  is. customary 
to  ignore  this  foreshortening,  and  make  all  the  isometrical  lines 
of  the  object  equal  to  their  true  lengths.  This  will  give  what 
is  called  the  isometrical  drawing  of  the  object,  which  will  be 
somewhat  larger  than  the  isometrical  projection. 

Fig.  65  represents  the  isometrical  projection  of  the  cube 
shown  in  Fig.  63,  and  Fig.  67  is  the  isometrical  drawing  of  the 
same  cube. 

115.  Definitions.  <?,  Fig.  65,  is  the  isometric  centre. 
cvbv,  cvdv,  and  cvev  are  the  isometric  axes.  Lines  parallel  to  either 
of  the  isometric  axes  are  isometric  lines,  while  any  line  not 
parallel  to  one  of  these  axes  is  a  non-isometric  line.  Planes 
parallel  to  the  faces  of  the  cube  are  isometric  planes,  and  those 
which  are  not  parallel  to  one  of  these  faces  are  non-isometric 
planes. 

116.  Direction  of  Light.  In  isometrical  drawing  the 
light  is  supposed  to  be  parallel  to  a  diagonal  of  the  cube,  as  in 
ordinary  projections,  only  here  it  is  parallel  to  the  plane  of 
projection,  and  it  is  represented  by  a  line  making  an  angle  of 
30°  with  a  horizontal  line.  Any  line  parallel  to  df,  Fig.  67, 
may  represent  a  ray  of  light. 

117.  Shade  Lines.  These  are  the  same  as  in  ordinary 
projections,  that  is,  they  separate  light  from  dark  surfaces.  In 
all   rectangular   objects  the  top,  left  front,  and  left  back  sur- 


114  ISOMETRICAL   DRAWING.  % 

faces  are  light,  while  the  bottom,  right  front,  and  right  back 
surfaces  are  dark.  In  Fig.  67  the  edges  ab,  be,  ce,  and  el  are 
the  visible  shade  lines.  And  all  rectangular  objects  have  their 
shade  lines  in  relative  positions  to  those  of  the  cube.  As  in 
ordinary  projections,  in  putting  shade  lines  on  a  group  of  ob- 
jects touchijig  each  other,  the  group  is  shaded  as  if  it  were  one 
solid ;  also,  in  outline  drawing  the  shadows  are  disregarded  in 
putting  in  shade  lines. 

118.  Pkob.  28.  To  make  the  isometrical  drawing  of  a  cube. 
Fig.  G7. 

With  the  centre  c  and  radius  equal  to  the  edge  of  the  cube 
draw  a  circle,  and  in  it  inscribe  a  regular  hexagon,  and  draw  the 
alternate  radii  cb,  cd,  and  ce,  and  the  drawiug  is  completed. 

Another  method,  which  is  applicable  to  any  rectangular 
object,  is  to  draw  from  any  point  as  e  lines  efeind  el,  each  mak- 
ing an  angle  of  30°  with  a  horizontal,  and  the  vertical  line  ce; 
on  these  lay  off  ef,  el,  and  ec  equal  to  the  true  length  of  the  edge 
of  the  cube ;  from  the  points/* and  I  draw  indefinite  vertical  Hues ; 
from  c  draw  the  lines  cb  and  cd  parallel  to  ef  and  el,  intersect- 
ing the  verticals  through/  and  I  in  the  points  b  and  d;  from 
the  points  b  and  d  draw  lines  parallel  to  cd  and  cb,  meeting  in 
a.     This  completes  the  drawing. 

119.  Prob.  29.  To  make  the  isometrical  drawing  of  a  rect- 
angular block,  with  another  rectangular  block  resting  on  its  top 
face,  and  a  recess  in  its  right  front  face.     Fig.  66. 

Construct  the  isometrical  drawing  of  the  large  block  by  the 
second  method  of  Art.  118. 

To  draw  the  small  block  it  is  first  necessary  to  locate  one  of 
its  lower  corners  in  the  top  face  of  the  large  one.  This  must 
be  done  by  means  of  two  coordinates  referred  to  two  isometric 


ISOMETRICAL   DRAWING.  115 

lines  as  axes.  The  point  e,  in  the  top  face  of  the  large  block, 
is  -J"  from  the  side  and  T3^"  from  the  end,  therefore  make  of 
equal  to  -J"  and  ag  equal  to  T3^",  then  from  f  and  g  draw  the 
isometric  lines  ef  and  ge,  intersecting  in  e,  the  point  required. 
The  rest  of  the  small  block  is  drawn  in  the  same  way  as  the 
large  one. 

To  make  the  recess  in  the  front  side  the  point  t  is  located 
in  the  same  way  as  the  point  e,  and  tx,  equal  to  the  depth  of 
the  recess,  is  laid  off  as  shown.  The  rest  of  the  construction 
is  evident. 

120.  Fig.  68  shows  the  isometrical  drawing  of  a  rectangular 
box,  without  cover,  15"  long,  6"  wide,  and  4"  high,  outside 
measurements,  the  boards  being  f"  thick.  The  scale  being 
1|-"  =  1'.  The  ends  are  nailed  on  to  the  sides,  and  the  bottom 
is  nailed  to  the  sides  and  ends.  The  visible  joints  are  shown. 
The  dotted  lines  show  the  inner  edges  of  the  box  which  are 
not  visible. 

121.  Fig.  69  shows  the  isometrical  drawing  of  a  four-armed 
cross.  A  careful  examination  of  the  figure  will  enable  the 
student  to  understand  its  construction,  there  being  only  iso- 
metric lines  involved. 

122.  Prob.  30.  To  make  the  isometrical  drawing  of  the 
pentagonal  prism  shown  in  Fig.  20.     Fig.  70. 

The  edges  of  the  base,  not  being  at  right  angles  to  each  other, 
are  non-isometric  lines,  hence  the  base  should  first  be  inscribed 
in  a  rectangle.  Let  one  side  of  the  rectangle  contain  the  edge 
be,  and  the  other  sides  respectively  contain  the  corners  a,  d, 
and  e.  Make  the  isometrical  drawing  of  the  rectangle  and 
locate  each  corner  of  the  base,  a,  b,  c,  d,  and  e,  by  laying  off  on 
the  sides  of  this  rectangle  the  distance  of  each   point  respect- 


116  ISOMETRIC AL  DRAWING. 

ively  from  the  nearest  corner  of  the  circumscribed  rectangle. 
At  these  points  draw  vertical  lines,  and  lay  off  on  each  of  them 
the  true  height  of  the  prism  and  join  the  tops,  completing  the 
drawing  of  the  prism.  It  will  be  noticed  that  the  non-isometric 
lines  on  the  drawing  are  not  equal  to  their  true  length  on  the 
object. 

123.  Prob.  31.  To  make  the  isometrical  drawing  of  an 
oblique  timber  framed  into  a  horizontal  one,  as  given  in  Fig.  53. 
Fig.  71. 

The  horizontal  timber  is  drawn  as  usual.  To  draw  the 
oblique  one,  these  edges  being  non-isometric,  two  points  have 
to  be  located  by  means  of  coordinates.  The  point  a  is  located 
by  making  ad  equal  to  the  distance  the  lower  end  of  the  oblique 
stick  is  from  the  end  of  the  horizontal  one.  Any  other  point 
b  is  found  by  making  cd  equal  to  the  horizontal  distance  of  the 
point  b  from  d,  and  be  equal  to  its  vertical  distance  from  d.  Join 
ab,  which  gives  the  isometrical  drawing  of  one  edge  of  the  ob- 
lique timber.  The  other  edges  are,  of  course,  parallel  to  this, 
being  drawn  through  the  points  e  and  f  which  are  located  the 
same  as  point  e  in  Prob.  30. 

124.  Prob.  32.  To  make  the  isometrical  drawing  of  the 
skeleton  frame  of  a  box  made  in  the  form  of  the  frustum  of  a 
square  pyramid.     Fig.  72. 

Let  a  square  prism  be  circumscribed  about  the  frustum. 
The  isometric  of  this  prism  is  readily  drawn,  and  is  shown  by 
the  dotted  lines.  The  bottom  edge  of  the  frustum  coincides 
with  the  bottom  of  the  prism.  The  points  a,  b,  c,  and  d  are  in 
the  upper  face  of  the  prism,  and  are  found  as  the  point  e  is  in 
Prob.  30. 

Join  ab,  be,  cd,  da,  af,  bg,  and  de,  and  the   main   frustum  is 


ISOMETRICAL   DRAWING.  117 

completed.     The  other  lines  which  change  the  frustum  from  a 
solid  to  a  skeleton  need  no  explanation. 

125.  Prob.  33.  To  make  an  exact  isometrical  drawing  of 
a  circular  card,  and  also  of  any  scroll  or  letter  on  its  surface. 
Fig.  74. 

Let  the  circle  and  letter  G  be  given  as  in  Fig.  73.  First, 
circumscribe  a  square  about  the  circle.  Make  the  isometric 
drawing  of  the  square.  The  centres  of  the  sides  of  the  square, 
d,  e,  f  and  g,  give  four  points  of  the  isometric  of  the  circle. 
Each  point  on  the  circumference  of  the  circle,  as  a,  b,  c,  etc., 
has  two  coordinates,  by  means  of  which  the  isometrical  draw- 
ing of  the  points  may  be  easily  obtained.  There  will  be  four 
points  on  the  circle  whose  coordinates  will  be  the  same  as  those 
for  c,  and  eight  points  whose  coordinates  will  be  the  same  as 
those  for  a  or  b,  or  any  point  between  d  and  e.  The  more 
points  that  are  taken  the  more  accurate  will  be  the  ellipse  which 
forms  the  isometrical  drawing  of  a  circle. 

The  letter  is  drawn  in  the  same  way  by  taking  the  two  coor- 
dinates of  any  point  on  it. 

126.  Prob.  34.  To  make  an  approximate  construction  of 
the  isometrical  drawing  of  a  circle.     Fig.  75. 

Make  the  isometrical  drawing  of  the  circumscribed  square  as 
before;  d,  e,f  and  g  will  be  four  points.  Draw  the  lines  ag, 
of,  bd,  and  be,  intersecting  in  the  points  c  and  o  ;  c  will  be  the 
centre  of  the  arc  between  d  and  g  ;  o  of  that  between  e  and/; 
a  of  that  between  g  and/;  and  b  of  that  between  d  and  e. 

127.  Fig.  7G  shows  the  approximate  construction  of  the 
isometrical  drawing  of  circles  in  each  of  the  three  visible  faces 
of  a  cube.     No  explanation  is  necessary. 


118  ISOMETRICAL   DRAWING. 

The  student  should  study  this  and  Fig.  75,  so  as  to  he  able 
to  make  the  isometrical  drawing  of  a  quarter  of  a  circle  in 
either  isometric  plane  without  making  the  whole  circle. 

128.  Fig.  78  shows  the  isometrical  drawing  of  a  bolt,  an 
hexagonal  nut,  and  a  circular  washer  as  shown  in  Fig.  77. 

129.  Fig.  79  shows  the  isometrical  drawing  of  the  hollow 
cylinder  given  as  in  Fig.  27. 

130.  Prob.  35.  To  divide  the  isometrical  drawing  of  a 
circle  into  equal  parts.     Fig.  74. 

At  the  middle  point  f  of  one  of  the  sides  nl  of  the  isomet- 
rical drawing  of  the  circumscribed  square  draw  fo  perpendicu- 
lar to  nl,  and  make  fo  equal  to  the  radius  of  the  circle  ;  draw 
ol  and  on ;  with  o  as  a  centre  and  radius  fo  describe  the  arc 
rfs  ;  divide  this  arc  into  any  number  of  parts  ;  draw  through  o 
and  these  points  of  division  lines  to  meet  nl;  join  these  meet- 
ing points  with  t,  the  centre  of  the  ellipse,  and  where  these  lines 
intersect  the  ellipse  will  be  the  points  of  division  of  the  iso- 
metrical drawing  of  one  quarter  of  the  circle. 

The  other  quadrants  can  be  divided  in  the  same  way. 

Another  Method.  On  the  long  diameter  of  the  ellipse 
draw  the  semicircle  ckp ;  divide  this  into  any  number  of  equal 
parts  ;  through  these  points  of  division  draw  lines  perpendicu- 
lar to  cp  ;  where  these  lines  intersect  the  ellipse  will  be  the 
points  of  division  sought. 

131.  In  isometrical  drawing  the  shadow  of  a  point  on  a  plane 
is  where  the  ray  of  light  through  the  point  intersects  its  projec- 
tion on  that  plane. 

To  find  the  shadow  of  the  line  dn  on  the  top  of  the  cube. 
Fig.  67. 

A  ray  of  light  through  the  point  n  is  nn3,  its  projection  on 


ISOMETRICAL   DRAWING.  119 

the  top  of  the  cube  is  dns  ;  these  two  lines  intersect  at  the  point 
ns,  which  is  the  shadow  of  the  point  n  on  the  top  of  the  cube. 
Join  this  point  with  d,  and  dns  is  the  shadow  of  the  vertical  line 
dn  required. 

To  find  the  shadow  of  the  line  do  on  the  left,  front  face  of  the 
cube. 

A  ray  of  light  through  the  point  o  is  oos,  its  projection  on  the 
face  of  the  cube  is  dos ;  these  lines  intersect  at  the  point  os, 
which  is  the  shadow  of  the  point  on  the  left,  front  face  of  the 
cube.    Join  this  with  d,  aud  dos  is  the  shadow  of  the  line  required. 

132.  Prob.  36.  To  find  the  shadow  of  a  cube  on  the  plane 
of  its  base.     Fig.  67. 

The  shadow  of  the  edge  ec  is  ecs ;  of  the  back  edge  ac  is 
cas  ;  of  the  point  b  is  bs ;  therefore,  the  shadows  of  ab  and  be 
are  asbs  and  bscs  respectively,  which  completes  the  shadow  re- 
quired. 

133.  In  Figs.  71  and  78  the  shadows  cast  by  the  objects  on 
each  other  and  on  the  ground  are  shown. 

134.  To  find  the  shadow  of  any  point  on  any  horizontal 
isometric  plane  proceed  as  follows :  Draw  though  the  point  a 
vertical  line,  and  make  it  equal  in  length  (downward  from  the 
point)  to  the  height  of  the  point  above  the  plane  receiving  the 
shadow ;  through  the  upper  end  of  this  line  draw  a  line  at  30° 
to  a  horizontal  in  the  direction  of  the  ray  of  light ;  through  the 
lower  end  draw  a  horizontal  line  ;  where  these  two  last  lines 
intersect  will  be  the  shadow  of  the  point. 

To  find  the  shadow  of  a  point  on  any  vertical  isometric 
plane  :  Draw  through  the  point  a  line  at  30°  to  a  horizontal, 
backward  and  to  the  right,  and  make  it  equal  in  length  (back- 
ward from  the  point)  to  the  perpendicular  distance  of  the  point 


120  ISOMETRICAL   DRAWING.     . 

from  the  plane ;  through  the  front  end  of  this  line  draw  a  line 
parallel  to  the  ray  of  light ;  through  the  back  end  draw  a  line 
at  60°  to  a  horizontal,  forward  and  to  the  right ;  where  these 
two  last  lines  intersect  will  be  the  shadow  of  the  point. 

OBLIQUE    PROJECTIONS. 

135.  Oblique  projections  differ  from  isometrical  projections 
onty  in  the  position  of  the  principal  faces  of  the  object.  In- 
stead of  being  placed  so  that  its  principal  faces  make  equal 
angles  with  the  plane  on  which  it  is  represented,  one  of  them 
is  placed  parallel  to  this  plane,  while  the  edges  which  represent 
the  remaining  dimension  of  the  object  may  be  drawn  at  any 
angle  with  a  horizontal,  for  convenience  usually  at  30°  or  45°. 
With  this  difference,  all  the  statements  and  principles  of  iso- 
metrical drawing  are  equally  applicable  to  oblique  projections. 

136.  Fig.  80  shows  the  oblique  projection  of  a  cube,  the 
shadow  of  lines  on  its  top  and  front  faces,  the  shadow  of  the 
cube  on  the  plane  of  its  base,  and  the  manner  of  drawing  the 
oblique  projection  of  circles  in  the  three  faces  of  the  cube. 
The  ellipse  in  the  top  face  is  drawn  by  the  arcs  of  circles, 
and  that  in  the  right  face  by  points  located  exactly  by  coordi- 
nates. 

It  will  be  noticed  that  the  ray  of  light  is  parallel  to  the  diag- 
onal df  of  the  cube,  and  that  its  projection  on  a  principal  sur- 
face is  parallel  to  the  diagonal  db  or  de. 

137.  Fig.  81  shows  the  oblique  projection  of  the  hollow 
cylinder  given  as  in  Fig.  27,  of  which  Fig.  79  is  the  isometri- 
cal drawing.  Figs.  1,  35,  and  36  are  oblique  projections  of 
models  representing  the  principles  of  projections  and  shadows. 


CHAPTER  VII. 


WORKING    DRAWINGS. 


138.  A  working  drawing  is  one  which  shows  all  the  dimen- 
sions of  an  object  in  such  a  way  that  the  object  could  be  repro- 
duced or  constructed  from  the  drawing.  Two  views  at  least, 
plan  or  horizontal  projection,  and  elevation  or  vertical  projec- 
tion, are  necessary ;  but  more  frequently  a  third  view,  usually 
an  end  view,  is  also  necessary.  Besides  these  views,  one  or 
more  sections  depending  upon  the  object  are  sometimes  neces- 
sary to  completely  determine  all  of  its  dimensions. 

139.  If  an  object  is  cut  through,  in  any  direction,  by  an 
imaginary  plane,  the  projection  of  one  part  of  the  object  on  a 
plane  parallel  to  the  cutting  plane,  the  person  supposed  to  be 
facing  this  cutting  plane,  is  called  a  section.  Sections  are  taken 
to  show  the  form  and  dimensions  of  the  interior  of  a  hollow 
object,  and  also  of  some  parts  of  solid  objects  which  are  not 
completely  determined  by  its  plan  and  elevations.  These  im- 
aginary cutting  planes  are  usually  taken  either  vertical  or  hori- 
zontal, but  it  is  sometimes  necessary  to  take  them  in  other 
positions,  but  perpendicular  to  a  vertical  or  horizontal  plane. 
All  that  part  of  the  object  which  is  cut  by  these  imaginary 
planes  is  cross-hatched,  that  is,  is  covered  by  parallel  lines  quite 
near  together.  The  direction  of  these  lines  should  be  either 
30°,  45°,  or  60°,  and  from  ■£./  to  £"  apart,  depending  on  the  size 
of  the  surface  to  be  cross-hatched,  the  smaller  the  nearer  to- 
gether.    They  may  be  drawn  in  either  direction. 


122  WORKING  DRAWINGS. 

140.  Fig.  82  represents  the  elevation,  Fig.  83  the  plan,  and 
Fig.  84  a  vertical  section  through  the  centre  on  the  line  AB, 
of  a  stuffing-box  gland.  The  side  elevation  was  not  necessary 
in  this  case,  as  all  the  dimensions  of  the  solid  are  shown  with- 
out it.     These  figures  are  made  one-half  size. 

141.  It  is  not  sufficient  simply  to  draw  the  projections  of  the 
object  the  correct  size,  but  the  dimensions  of  the  solid  should 
all  be  clearly  placed  on  the  drawing,  so  that  the  workman,  or 
whoever  has  occasion  to  read  the  drawing,  is  not  obliged  to 
use  the  scale,  thereby  removing  a  great  liability  to  error.  These 
dimensions  should  be  put  in  neatly,  and  should  follow  a  certain 
system.  The  method  used  in  Figs.  82,  83,  and  84  should  be 
carefully  observed,  besides  which  the  following  general  direc- 
tions for  putting  in  dimensions  and  representing  special  feat- 
ures should  be  followed  : 

In  placing  dimensions  upon  the  drawing  a  line  should  be 
drawn  from  one  point  to  another,  between  which  the  dimension' 
is  to  be  given,  and  the  actual  dimension,  or  distance  apart  of 
the  points,  is  placed  in  the  line,  a  space  having  been  left  for  it 
near  the  centre.  These  lines  should  be  fine  and  composed  of 
dashes  about  %"  long  with  about  £'  spaces  between  them.  When 
the  dimension  is  small  of  course  the  dashes  must  be  made 
shorter.  Arrow  heads  are  placed  at  the  ends  of  these  lines, 
the  point  of  the  arrow  exactly  touching  the  points  or  lines  be- 
tween which  the  dimension  is  given,  the  arrow  heads  pointing 
away  from  each  other.  When  the  dimension  is  very  small  the 
arrow  heads  may  be  placed  on  the  outside  of  the  lines  instead 
of  between  them,  and  in  that  case  should  point  toward  each 
other. 

The  arrow  heads  should  be  drawn  free-hand  and  not  made 
with  the  drawing  pen.  The  figures  for  dimensions  should  also 
be  made  free-hand,  and  should  always  be  placed  at  right  angles 


WORKING   DRAWINGS.  123 

to  the  dimension  line,  and  should  read  from  the  hottom  or  right- 
hand  side  of  the  drawing.  They  should  be  put  down  in  inches 
and  thirty-seconds,  sixteenths,  eights,  quarters,  and  halves,  as 
the  case  may  be,  thus,  1T5^' ',  not  f£".  The  fractions  should  be 
reduced  to  their  lowest  terms,  thus,  •§-" ,  not  -J§".  The  dividing 
line  in  the  fraction  should  always  be  made  horizontal,  as  they 
are  less  likely  to  be  misunderstood.  The  inch  marks  should 
be  placed  after  the  fraction. 

On  rough  castings  measure  to  the  nearest  sixteenth,  on  ordi- 
nary fiuished  surfaces  take  the  nearest  thirty-second,  and  on 
fine  finish  and  fits  be  as  accurate  as  possible. 

All  dimensions  up  to  two  feet  should  be  put  down  in  inches, 
thus,  15",  22f",  and  all  above  that  in  feet  and  inches,  thus, 
3'-2",  or  3  ft.  2".  Students  should  be  very  careful  to  get  all 
of  the  important  dimensions  on,  and  also  an  "  overall "  dimen- 
sion, so  that  the  workman  will  not  be  required  to  add  a  num- 
ber of  dimensions  together.  Important  dimensions  are  those 
which  are  necessary  for  the  workman  to  construct  the  piece. 
The  dimensions  should  not  interfere  with  each  other,  and  care 
should  be  taken  not  to  have  them  cross  each  other  in  a  circle. 

As  a  general  thing  do  not  repeat  any  dimensions  ;  that  is,  if 
a  dimension  is  given  on  one  view  do  not  repeat  it  on  another. 

In  order  that  the  drawing  may  be  left  as  distinct  as  possible 
it  is  frequently  advisable  to  put  the  dimensions  outside  the  fig- 
ure, or  better,  where  two  or  more  views  are  given,  put  them 
between  the  different  views,  as  shown  in  Figs.  82,  83,  and  84  ; 
the  widths  being  placed  between  the  plan  and  elevation,  and 
the  heights  between  the  elevation  and  section,  or  between  the 
two  elevations  where  an  end  elevation  is  given.  To  do  this 
"  extension  "  lines  must  be  used.  They  should  be  composed  of 
fine  dash  lines,  the  dashes  being  about  £"  long,  so  as  to  be  dis- 
tinguished easily  from  the  dotted  lines  representing  the  invisi- 


124  WORKING  DRAWINGS. 

ble  parts  of  an  object.  Where  the  dimensions  do  not  interfere 
with  the  drawing,  as  is  the  case  in  Fig.  84,  it  is  better  to  put 
them  on  the  figure  between  the  lines  themselves,  or  as  near  as 
possible. 

Give  the  diameter  of  a  circle  instead  of  the  radius.  When 
only  an  arc  is  shown  give  the  radius,  and  draw  a  very  small 
circle  about  its  centre,  and  let  this  circle  take  the  place  of  an 
arrow  head.  The  dimension  line  should  be  drawn  from  the 
edge  of  this  circle  and  not  from  its  centre. 

In  locating  holes  or  bolts  the  dimensions  should  be  given 
from  the  outside  of  the  piece  to  the  centre  of  the  hole  or  bolt, 
and  their  distance  apart  is  shown  by  giving  the  distance  be- 
tween centres.  See  Fig.  85.  Holes  are  very  often  located 
from  the  centre  line  of  a  piece,  so  that  it  is  unnecessary  to  give 
the  dimension  from  the  outside  of  the  piece. 

The  "  centre  "  line  should  be  composed  of  long  and  short 
dashes,  the  long  dashes  being  about  £"  long  and  the  short  ones 
about  -J"  long. 

If  the  holes  are  arranged  in  a  circle,  as  in  Fig.  86,  give  the 
diameter  of  the  circle  passing  through  the  centre  of  the  holes. 

In  drawing  a  bolt  or  screw  represent  the  threads  as  shown  in 
Fig.  87  ;  it  is  not  necessary  that  the  spaces  should  correspond 
with  the  true  pitch  of  the  threads.  In  order  to  obtain  the  cor- 
rect slant  of  the  threads,  a  line  drawn  at  right  angles  to  the 
axis  of  the  bolt  should  pass  through  the  point  of  a  thread  on 
one  side,  and  the  centre  of  a  space  on  the  other. 

Always  give  that  view  of  a  square,  hexagonal,  or-  octagonal 
bolt-head,  or  nut  which  shows  the  distance  between  its  paral- 
lel sides. 

In  placing  the  dimensions  upon  a  bolt  or  screw,  always  give 
the  length  of  the  unthreaded  part  in  addition  to  the  length  of 
the  bolt.  The  length  of  the  bolt  should  be  given  from  the 
under  side  of  the  head  to  the  extreme  end.     See  Fig.  87. 


WOEKING  DEAWINGS.  125 

In  making  a  sectional  view  on  a  line  passing  lengthwise 
through  the  centre  of  a  shaft,  bolt,  or  screw,  it  is  generally 
unnecessary  to  represent  the  shaft,  bolt,  or  screw  in  section,  as 
the  view  is  more  clearly  shown  by  leaving  them  in  full.  See 
Fig.  89. 

In  drawings  where  bolts  or  screws  are  shown  by  clotted  lines 
do  not  dot  in  the  threads,  but  represent  them  by  double  dotted 
lines,  as  shown  in  Fig.  89. 

Represent  a  tapped  hole  as  shown  in  Fig.  90. 

The  line  on  which  a  section  is  taken,  as  AB,  Fig.  83,  should 
be  made  the  same  as  a  centre  line. 

142.  Although  it  is  customary  to  represent  a  screw  by 
straight  lines,  as  shown  in  Fig.  87,  it  is  sometimes  desirable  to 
make  its  actual  projections,  especially  if  the  screw  be  a  large 
one. 

The  thread  of  a  screw  is  a  curve  which  is  called  a  helix.  A 
cylindrical  helix  is  generated  by  a  point  caused  to  travel  round 
a  cylinder,  having,  at  the  same  time,  a  motion  in  the  direction 
of  the  length  of  the  cylinder, —  this  longitudinal  motion  bear- 
ing some  regular  prescribed  proportion  to  the  circular  or  angu- 
lar motion.  The  distance  between  any  two  points  which  are 
nearest  to  each  other,  and  in  the  same  straight  line  parallel  to 
the  axis  of  the  cylinder,  is  called  the  pitch, —  in  other  words, 
the  longitudinal  distance  traversed  by  the  generating  point  dur- 
ing one  revolution. 

To  draw  the  projections  of  a  helix.     Fig.  91. 

The  plan  of  the  helix  will  be  a  circle.  Divide  this  circle 
into  any  number  of  equal  parts,  in  this  case  twelve  ;  divide  the 
pitch  into  the  same  number  of  equal  parts.  It  is  evident  that 
when  the  point  has  moved  one-twelfth  the  distance  around  the 
circumference,  it  has  also  moved  in  the  direction  of  the  axis 
one-twelfth  of  the  pitch ;   when  it  has  moved  two-twelfths  the 


126  WORKING  DRAWINGS. 

distance  around  it  has  moved  two-twelfths  of  the  pitch  ;  there- 
fore, from  the  points  of  division,  a\  bh,  ch,  etc.,  in  the  plan  draw 
vertical  lines  until  they  intersect  horizontal  lines  drawn  from 
the  corresponding  division  of  the  pitch.  And  these  intersec- 
tions, av,  bv,  cv,  etc.,  will  be  points  on  the  vertical  projection  of 
the  helix. 

Fig.  92  shows  a  V-threaded  screw  in  projection. 


CHAPTER  VIII. 


EXAMPLES. 


All  polygons  referred  to  in  these  Examples  are  regular  polygons,  unless  other- 
wise stated. 

1.  Draw  the  two  projections  of  a  point  1^"  from  H  and  l" 
from  V. 

2.  Of  a  point  lying  in  H  and  f "  in  front  of  V. 

3.  Of  a  point  lying  in  V  and  l"  above  H. 

4.  Of  a  line  1"  long,  parallel  to  both  V  and  H,  f"  above  H 
and  1"  in  front  of  V. 

5.  Of  same  line  when  it  is  perpendicular  to  V  and  14/' 
above  H,  its  back  end  being  4/'  in  front  of  V. 

6.  Of  same  line  when  it  is  perpendicular  to  H  and  1"  in 
front  of  V,  its  lower  end  being  4/'  above  H. 

7.  Of  same  line  when  it  is  parallel  to  H,'  1"  above  H  and 
making  an  angle  of  30°  with  V,  its  back  end  being  \"  in  frout 
of  V. 

8.  Of  same  line  when  it  is  parallel  to  V,  f "  in  front  of  V 
and  making  an  angle  of  60°  with  H,  its  lower  end  being  4/' 
above  H. 

9.  Of  same  line  lying  in  H  and  making  an  angle  of  45° 
with  V,  its  back  end  being  4/'  in  front  of  V. 

10.  Of  same  line  lying  in  V,  parallel  to  and  1"  above  H. 

11.  Of  same  line  when  it  is  inclined  at  an  angle  of  60°  with 
H,  and  whose  horizontal  projection  makes  an  angle  of  45°  with 
GL,  one  end  being  4/'  above  H  and  4/'  in  front  of  V. 


128  EXAMPLES. 

12.  A  wire  1-j"  long  projects  from  a  vertical  wall  at  60° 
with  the  surface,  and  is  parallel  to  the  ground  and  1"  above  it. 
Draw  plan  and  elevation. 

13.  Find  true  length  of  a  line  given  by  its  projections,  as 
follows  :  One  end  is  \"  from  each  plane  and  the  other  is  2" 
above  H,  the  horizontal  projection  of  the  line  is  1£"  long  and 
makes  an  angle  of  30°  with  GL. 

14.  Draw  plan  and  elevation  of  a  line  1^-"  long,  lying  in  a 
profile  plane,  making  an  angle  of  60°  with  H,  whose  lower  end 
is  |-"  from  V  and  \"  from  H. 

15.  Of  an  oblique  line,  one  end  being  above  and  in  front  of 
the  other.     Find  its  true  length  and  angle  it  makes  with  H. 

16.  Of  an  oblique  line,  one  end  being  behind  and  above 
the  other.     Find  its  true  length  and  angle  it  makes  with  V. 

17.  Of  a  line  which  slopes  downward,  backward,  and  to 
the  right.     Find  its  true  length  by  revolving  parallel  to  V. 

18.  Of  a  line  which  slopes  downward,  forward,  and  to  the 
left.     Find  its  true  length  by  revolving  parallel  to  H. 

19.  Of  two  lines  which  are  parallel  in  space  and  slope 
downward,  forward,  and  to  the  right. 

20.  Of  a  line  2"  long,  sloping  downward,  forward,  and  to 
the  right,  one  end  being  \\"  above  H  and  w"  in  front  of  V,  the 
other  end  A/'  above  H  and  14/'  in  front  of  V. 

21.  Draw  plan  and  elevation  of  a  rectangular  card  f"  x  14/' 
which  is  perpendicular  to  H,  parallel  to  V.  and  §-"  in  front  of 
V ;  its  short  sides  are  parallel  to  H  and  the  lower  one  is  4/' 
above  H.  Revolve  this  card  forward  about  its  left-hand  edge 
(like  a  door  on  its  hinges)  through  angles  of  30°,  45°,  60°, 
and  90°,  and  construct  corresponding  plans  and  elevations. 

22.  Of  same  card  when  it  is  lying  on  H  with  its  long  sides 
parallel  to  V  and  4/'  in  front  of  V.  Revolve  card  about  right- 
hand  horizontal  edge   (like  a  trap-door  on  its  hinges)  through 


EXAMPLES.  129 

angles  of  30°,  45°,  60°,  and  90°,  and  construct  corresponding 
projections. 

23.  Of  same  card  when,  besides  making  the  angles  of  0°, 
30°,  45°,  60°,  and  90°  with  H,  as  in  last  example,  the  hori- 
zontal projection  of  its  long  edges,  in  all  its  different  positions, 
makes  an  angle  of  30°  with  GL  backward  and  to  the  right. 

24.  Of  same  card  when  it  is  parallel  to  V  and  4/'  in  front 
of  V,  one  of  its  diagonals  being  parallel  to  H.  Revolve  this 
card  through  an  angle  of  60°  about  a  vertical  axis,  and  construct 
its  corresponding  projections. 

25.  Of  same  card  when,  besides  making  an  angle  of  60° 
with  V,  as  in  last  example,  the  vertical  projection  of  the  diago- 
nal which  was  parallel  to  H  makes  an  angle  of  45°  with  GL. 

26.  Of  a  card  £"  square  resting  on  II,  with  one  diagonal 
parallel  to  V  and  1"  in  front  of  V,  then  raise  left-hand  end  of 
diagonal  until  it  makes  an  angle  of  45°  with  14  and  construct 
corresponding  projections. 

27.  Of  the  same  card  when,  besides  making  an  angle  of  45° 
with  H,  as  in  last  example,  the  horizontal  projection  of  the 
diagonal  makes  an  angle  of  30°  with  GL. 

28.  Same  as  example  27,  except  that  the  horizontal  projec- 
tion of  the  diagonal  makes  an  angle  of  90°  with  GL. 

29.  Of  an  hexagonal  card  whose  sides  are  §■"  long,  when 
one  of  its  long  diameters  is  parallel  to  V  and  1"  in  front  of  V, 
one  end  resting  on  H.  The  surface  of  the  card  is  perpendicu- 
lar to  V  and  makes  an  angle  of  30°  with  H. 

30.  Of  same  card  when  its  diameter,  besides  making  an 
angle  of  30°  with  H,  as  in  last  example,  has  its  horizontal  pro- 
jection inclined  at  an  angle  of  60°  with  GL. 

31.  Of  a  pentagonal  card  whose  surface  is  perpendicular  to 
H  and  makes  an  angle  of  45°  with  V,  one  edge  being  perpen- 
dicular to  H  and  resting  against  V.  The  diameter  of  the  cir- 
cumscribed circle  is  1^". 


130  EXAMPLES. 

32.  Of  same  card  when,  besides  making  an  angle  of  45° 
with  V,  as  in  last  example,  it  has  been  revolved  through  an 
angle  of  30°. 

33.  Of  an  octagonal  card  resting  on  one  of  its  edges,  with 
its  surface  perpendicular  to  V  and  inclined  at  an  angle  of  60° 
with  H.     The  diameter  of  the  inscribed  circle  is  1^". 

34.  Of  same  card  when  it  is  inclined  at  same  angle  with  H 
as  in  last  example,  and  the  horizontal  projection  of  the  edges 
which  were  perpendicular  to  V  makes  an  angle  of  45°  with 
GL. 

35.  Of  an  isosceles  triangle  situated  in  a  profile  plane,  the 
base  making  an  angle  of  15°  with  H,  its  back  corner  being  §■" 
above  H  and  £■"  in  front  of  V,  and  lower  than  its  front  corner. 
The  base  of  triangle  is  1^"  and  altitude  1". 

36.  Of  an  hexagonal  card  whose  surface  is  perpendicular 
to  both  V  and  H.  Its  long  diameter  is  1".  Two  of  its  edges 
are  perpendicular  to  H.  Centre  of  card  is  1"  above  H,  and 
1£"  in  front  of  V. 

37.  Of  a  circular  card  3"  in  diameter,  perpendicular  to  H 
and  making  an  angle  of  30°  with  V,  resting  on  H,  back  edge 
1"  from  V. 

38.  Of  same  card  after  it  has  been  revolved  through  an 
angle  of  45°,  keeping  the  same  angle  with  V  as  in  last  exam- 
ple. 

39.  Of  a  circular  card  3"  in  diameter,  perpendicular  to  H 
and  making  an  angle  of  45°  with  V.  Solve  as  explained  in 
Art.  61. 

40.  Draw  plan  and  elevation  of  a  cube  of  1"  edge  resting 
on  H,  £"  in  front  of  V,  with  two  faces  parallel  to  V.  Find 
the  true  size  of  the  angle  which  the  diagonal,  which  slopes 
downward,  backward,  and  to  the  right,  makes  with  V  and  also 
with  H. 


EXAMPLES.  131 

41.*  Of  a  rectangular  prism  whose  base  is  f"  x  1"  and 
length  is  1-^-",  resting  with  its  base  against  V,  its  lower  left-hand 
face  making  an  angle  of  60°  with  V. 

Shade  lines  are  to  be  put  in  in  all  the  examples  where  there  are  any. 

42.  Of  a  cylinder  resting  with  its  base  on  H ;  diameter  of 
cylinder  is  1"  and  its  length  is  1^".  The  axis  is  f"  in  front 
of  V. 

43.  Of  a  cone  resting  with  its  base  parallel  to  and  £■"  in 
front  of  V.     The  diameter  of  base  is  14/'  and  its  height  is  If". 

44.  Of  a  heptagonal  prism  resting  with  its  base  on  H,  one 
of  its  faces  making  an  angle  of  15°  with  V.  The  diameter  of 
the  circumscribed  circle  about  base  is  1"  and  the  height  of 
prism  is  If". 

45.  Of  an  octagonal  pyramid  resting  with  its  base  on  H, 
with  two  of  the  edges  of  the  base  making  an  angle  of  30°  with 
GL.  The  diameter  of  the  circle  inscribed  in  the  base  is  l-§-" 
and  the  altitude  is  1". 

46.  Draw  plan  and  two  elevations  of  a  square  prism  with 
its  axis  parallel  to  both  V  and  H ;  the  axis  is  1"  above  H  and 
14/'  in  front  of  V,  base  of  prism  is  £"  square,  and  the  length  is 
14/' ;  the  upper  left-hand  long  face  makes  an  angle  of  30° 
with  H. 

47.  Draw  plan  and  elevation  of  same  prism  when  its  axis 
is  parallel  to  H  and  makes  an  angle  of  45°  with  V,  backward 
and  to  the  left. 

48.  Draw  plan  and  two  elevations  of  an  hexagonal  prism 
2"  long,  diameter  of  circumscribed  circle  about  base  is  14/', 
the  axis  is  parallel  to  H  and  makes  an  angle  of  30°  with  V, 
backward  and  to  the  right ;  lower  edge  of  prism  rests  on  II, 
and  the  lower  right-hand  face  makes  an  angle  of  20°  with  H. 

*  The  base  is  supposed  to  be  at  right  angles  to  the  axis  in  all  the  prisms,  pyra- 
mids, cylinders,  and  cones,  unless  otherwise  stated. 


132  EXAMPLES. 

49.  Draw  plan  and  two  elevations  of  a  circular  cylinder, 
3"  in  diameter  and  2|"  long,  with  a  circular  hole  through  it 
ljl"  in  diameter  ;  axis  is  parallel  to  both  V  and  H. 

50.  Draw  plan  and  elevation  of  same  cylinder  resting  with 
its  base  on  H. 

51.  Of  same  cylinder  when  lying  on  H,  with  its  axis  par- 
allel to  H  and  makiug  an  angle  of  60°  with  V. 

52.  Draw  plan  and  two  elevations  of  a  pile  of  blocks 
located  as  follows :  the  lowest  one  is  3"  long.  14/'  wide,  and  4/' 
thick,  it  rests  with  its  wide  face  on  H,  its  long  edge  making 
an  angle  of  30°  with  V,  backward  and  to  the  right ;  on  top  of 
this  a  second  block  rests,  equal  in  width  and  thickness  to  the 
first  but  1"  shorter  ;  on  this  a  third  block  rests,  of  the  same 
width  and  thickness  as  the  others,  but  1"  shorter  than  the  sec- 
ond ;  these  blocks  are  placed  symmetrically. 

53.  Draw  plan  and  elevation  of  a  pyramid  formed  of  four 
equilateral  triangles  of  2"  sides,  when  one  edge  of  the  base  is 
at  30°  with  V. 

54.  Of  an  hexagonal  prism  standing  with  its  base  on  H, 
two  of  its  faces  making  angles  of  20°  with  V ;  diameter  of 
circumscribed  circle  about  base  is  8",  length  of  prism  is  10", 
scale  3"  =  V,  or  ^  size. 

55.  Of  same  prism  when  axis  is  parallel  to  V  and  makes 
an  angle  of  60°  with  H  and  slopes  downward  to  the  left. 
Scale  3"=1'. 

56.  Of  same  prism  when  its  axis,  besides  making  an  angle 
of  60°  with  H,  has  its  horizontal  projection  inclined  at  an  angle 
of  60°  with  GL  ;  axis  of  prism  slopes  downward,  forward,  and 
to  the  left.     Scale  2"=1'. 

57.  Of  a  pentagonal  pyramid  resting  with  its  base  on  H, 
one  edge  of  base  perpendicular  to  V,  diameter  of  circumscribed 
circle  16",  height  of  pyramid  20".  Find  its  shadow.  Scale 
!■§■"= 1',  or  4,  size. 


EXAMPLES.  133 

58.  Of  same  pyramid  when  its  axis  is  parallel  to  V  and 
slopes  downward  to  the  left,  making  an  angle  of  75°  with  H. 
Find  its  shadow.     Scale  ].■£■"=  1'. 

59.  Of  same  pyramid  when  its  axis,  besides  making  an  angle 
of  75°  with  H,  has  its  horizontal  projection  inclined  at  an  angle 
of  30°  with  GL,  so  that  axis  slopes  downward,  backward,  and 
to  the  right.     Find  its  shadow.     Scale  L$-"  =  l'. 

60.  Of  a  cone  resting  with  its  base  on  H,  diameter  of  base 
20",  height  of  cone  2'.     Find  its  shadow.     Scale  l£"=l'. 

61.  Of  same  cone  when  resting  with  an  element  on  H, 
with  its  axis  parallel  to  V.     Find  its  shadow.     Scale   1-J-"*  =  1*. 

62.  Of  same  cone  with  an  element  on  H,  and  axis  making 
an  angle  of  45°  with  GL.     Find  its  shadow.     Scale  1|"  =  1'. 

63.  Of  same  cone  with  an  element  on  H,  and  axis  lying 
in  a  profile  plane,  sloping  downward  and  backward.  Find  its 
shadow.     Scale  1^"  =  1'. 

64.  Of  the  frustum  of  an  octagonal  pyramid  resting  with 
its  base  on  H,  long  diameter  of  lower  base  is  3"  and  of  the 
upper  base  is  2",  the  height  of  frustum  is  3",  the  front  left-hand 
edge  of  base  makes  an  angle  of  15°  with  GL,  backward  to  the 
left.  There  is  a  hole  1"  square  through  the  centre  of  frustum, 
whose  axis  is  coincident  with  axis  of  frustum ;  two  sides  of 
the  hole  make  angles  of  7^-°  with  GL. 

65.  Of  same  frustum  when  its  axis  is  parallel  to  V  and 
makes  an  angle  of  60°  with  H  and  slopes  downward  to  the  left. 

66.  Of  same  frustum  when  its  axis,  besides  making  an  angle 
of  60°  with  H,  has  its  horizontal  projection  inclined  at  an  angle 
of  30°  with  GL,  and  slopes  downward,  forward,  and  to  the  left. 

67.  Of  the  skeleton  frame  of  a  box  3'  long,  2'  wide,  and  2' 
high,  the  joists  being  all  2£"  square.  The  frame  rests  on  H 
with  its  long  sides  parallel  to  V.  Do  not  show  joints  in  framing. 
Scale  l"=r. 


134  EXAMPLES. 

68.  Of  same  frame  still  resting  on  H  with  its  long  sides 
making  an  angle  of  50°  with  V,  backward  to  the  left.     Scale 

r=i'. 

69.  Revolve  the  elevation  obtained  in  Example  68  through 
an  angle  of  30°  (in  either  direction),  and  construct  correspond- 
ing plan.     Scale  1"=1'. 

70.  Of  a  double  cross  standing  on  its  base,  one  arm  paral- 
lel to  both  V  and  H;  upright  piece  is  1'— 8"  square  and  10'— 8" 
high,  each  arm  is  l'-8"  square  and  7'  long  (out  to  out),  their 
top  surfaces  are  2'— 8"  below  the  top  of  upright.     Scale  §"  =  1'. 

71.  Of  same  cross  when  its  axis  is  parallel  to  V  and  makes 
an  angle  of  60°  with  H.     Scale  f"=l'. 

72.  Of  same  cross  when  its  axis,  besides  making  an  angle 
of  60°  with  H,  has  its  horizontal  projection  inclined  at  an 
angle  of  60°  with  H,  sloping  downward,  forward,  and  to  the 
left.     Scale  f-"=l\ 

73.  Of  a  pyramid  resting  on  its  apex  with  axis  perpendicu- 
lar to  H  and  2'-6"  in  front  of  V,  its  base  is  an  equilateral  tri- 
angle of  2-8''  sides  and  its  altitude  is  11".  The  left-hand  edge 
of  base  is  perpendicular  to  V.     Find  its  shadow.     Scale  f"  =  l'. 

74.  Of  same  pyramid  when  its  axis  is  parallel  to  V  and 
makes  an  angle  of  60°  with  H,  and  slopes  downward  to  the 
right.     Find  its  shadow.     Scale  f" =1'. 

75.  Of  same  pyramid  when  its  axis,  besides  making  an  angle 
of  60°  with  H,  has  its  horizontal  projection  inclined  at  an  angle 
of  30°  with  GL,  so  that  it  slopes  downward,  forward,  and  to 
the  left,  its  apex  being  1'— 6"  from  V,  still  resting  on  H.  Find 
its  shadow.     Scale  f"=l'. 

76.  There  is  a  solid  formed  of  two  equal  square  pyramids 
of  2"  base  and  3"  altitude,  which  are  united  by  their  bases. 
Draw  plan  and  elevation  when  the  object  rests  with  one  of  its 
triangular  faces  on  H,  its  axis  being  parallel  to  V  and  2£"  in 
front  of  V.     Find  its  shadow. 


EXAMPLES.  135 

77.  Of  same  object  still  resting  on  one  of  its  faces,  when 
the  horizontal  projection  of  the  axis  makes  an  angle  of  45° 
with  GL,  and  slopes  downward,  backward,  and  to  the  right. 
Find  its  shadow. 

78.  Draw  projections  of  a  pentagonal  prism  whose  length 
is  24/'  and  radius  of  circumscribed  circle  about  the  end  is  4/' ; 
the  prism  rests  with  one  of  its  long  edges  on  H,  which  makes 
an  angle  of  60°  with  V,  backward  to  the  left,  and  whose  front 
end  is  3"  from  V.  The  lower  left-hand  long  face  makes  an 
angle  of  15°  with  H. 

Also,  of  a  triangular  pyramid  resting  on  its  base  on  H,  with 
its  axis  If"  to  the  left  of  the  point  located  in  prism,  and  4"  in 
front  of  V,  diameter  of  circumscribed  circle  is  2",  the  altitude 
of  pyramid  is  34/',  right-hand  edge  of  base  is  perpendicular  to 
V.  Find  shadow  of  prism  on  H  and  V,  also  of  pyramid  on  H 
and  on  prism. 

79.  Find  the  shadow  on  PI  of  a  card  |"  square,  parallel  to 
H,  and  4/'  above  H,  two  edges  making  angles  of  30°  with  V. 

80.  Of  same  card  on  H,  when  it  is  parallel  to  V,  2"  in  front 
of  V,  two  edges  parallel  to  H,  lowest  edge  4/'  above  H. 

81.  Of  same  card  on  V  and  H,  lying  in  a  profile  plane,  two 
edges  perpendicular  to  H,  back  edge  4/'  in  front  of  V  and  low- 
est edge  4/'  above  H. 

..  82.  Of  an  hexagonal  card,  parallel  to  V,  4/'  in  front  of  V, 
two  edges  parallel  to  H,  centre  of  hexagon  14/'  above  H,  long 
diameter  1". 

83.  Of  same  card  parallel  to  H,  14/  above  H,  centre  1"  in 
front  of  V,  two  edges  parallel  to  V. 

84.  Of  a  circular  card,  14/'  in  diameter,  parallel  to  V,  14/' 
in  front  of  V,  centre  1"  above  -H. 

85.  Of  an  hexagonal  card  whose  surface  is  perpendicular 
to  V  and  H,  two  of  its  edges  perpendicular  to  H,  centre  of 


136  EXAMPLES. 

hexagon  \§"  above  II,  1-^"  in  front  of  V,  diameter  of  inscribed 
circle  1".  In  constructing  projections  of  hexagon  revolve  it 
about  a  vertical  axis  through  centre. 

86.  Of  a  cube  of  f"  sides,  parallel  to  V  and  H,  1£"  above 
H,  and  £"  in  front  of  V. 

87.  Of  a   square  prism  standing  on  H,  each  face  f-"  x  l£", 
two  faces  parallel  to  V,  -J"  in  front  of  V. 

88.  Of  same  prism  still  standing  on  H,  turned  so  that  two 
faces  make  angles  of  30°  with  V,  backward  to  the  right. 

89.  Of  a  cylinder  f"  in  diameter  and  If"  high,  with  base 
resting  against  V  1"  above  H. 

90.  Of  a  cone  1^"  high,  base  f-"  in  diameter,  standing  on  H, 
with  axis  If"  in  front  of  V. 

91.  Of  a  line  located  as  shown  in  Example  20. 

92.  Of  a  card  located  as  shown  in  Example  36. 


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